4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 33 InterviewSolution

1601.	From the Maxwell distribution function frind `lt lt v_x^2 gt gt`, the mean value of the squared `v_x` projection of the molecular velocity in a gas at a temperature `T`. The mass of each molecule is equal to `m`.
Answer» `lt v_x^2 gt = int_0^oo v_x^2 e - (mv_x^2)/(2 kt) dv_x//int_0^oo e - (mv_x^2)/(2 kt) dv_x` `=(2 kT)/(m) int_0^oo xe^-x (dx)/(2 sqrt(x))//int_0 ^oo e^-x (dx)/(2 sqrt(x))` =`(2 kT)/(m) Gamma((3)/(2))// Gamma ((1)/(2)) = (kT)/(m)`.

Discussion

1602.	Using the Maxwell distribution function, determine the pressure exterted by gas on a wall, if the gas temperature is `T` and the concentration of molecules is `n`.
Answer» Let, `dn(v_x) = n ((m)/(2 pi k T))^(1//2) e^(-mv_x^2//2 kT) dv_x` be the number of molecules per unit volume with `x` component of velocity in the range `v_x` to `v_x + dv_x` Then `p = int_0^oo 2mv_x.v_x dn(v_x)` =`int_0^oo 2 mv_x^2 n((m)/(2 pi kT))^(1//2) e^(-mv_x^2//2 kT) dv_x` =`2mn (1)/(sqrt(pi)) (2kT)/(m) int_0^oo u^2 e^(-u^2) du` =`(4)/(sqrt(pi)) nkT. int_0^oo x ^(-x) (dx)/(2 sqrt(x)) = nkT`.

Discussion

1603.	Find the mass of a mole of colloid particles if during their centrifuging with an angular velocity `omega` about a vertical axis the concentration of the particles at the distance `r_2` from the rotation axis is `eta` times greater than that at the distance `r_1` (in the same horizontal plane). The densities of the particles and the solvent are equal to `rho` and to `rho_0` respectively.
Answer» In a centrifuge rotating with angular velocity `omega` about an axis, there is a centrifugal acceleration `omega^2 r` where `r` is the radial distance from the axis. In a fluid if there are suspened colloidal particle they experience an additional force. If `m` is the mass of each particle then its volume is `(m)/(rho)` and the excess force on this particle is. `(m)/(rho) (rho - rho_0) omega^2 r` outward corresponding to a potential energy `-(m)/(2 rho)(rho - rho_0) omega^2 r^2` This gives rise to a concentration variation. `n(r) = n_0 exp(+ (m)/(2 rho kT)(rho - rho_0) omega^2 r^2)` Thus `(n(r_2))/(n(r_1)) = eta = exp (+(M)/(2 rho RT) (rho - rho_0) omega^2 (r_2^2 - r_1^2))` where `(m)/(k) = (M)/(R), M = N_A m` is the molecular weight Thus `M = (2 rho RT 1n eta)/((rho - rho_0)omega^2(r_2^2 - r_1^2))`.

Discussion

1604.	A very tall vertical cylinder contains a gas at a temperature true `T`. Assuming the gravitational field to be uniform, find the mean value of the potential energy of the gas molecules. Does this value depend on whether the gas consists of one kind of molecules or of serveral kinds ?
Answer» `lt U gt = (int_0^oo mgz e^(-mgz//kT) dz)/(int_0^oo e^(-mgz//kT dz)) = kT (int_0^oo x e^(-x) dx)/(int_0^oo e^-x dx) = kT (Gamma (2))/(Gamma(1)) = kT` When there many kinds of molecules. This formula holds for each kind and the average energy. `lt U gt = (sum f_i kT)/(sum f_i) = kT` Where `f_i alpha` fractional concentration of each kind at the ground level.

Discussion

1605.	A very tall vertical cylinder contains carbon dioxide at a certain temperature `T`. Assuming the gravitational field to be uniform, find how the gas pressure on the bottom of the vessel will change when the gas temperature increases `eta` times.
Answer» At a temperature `T` the concentration `n(Ƶ)` varies with height according to `n (Ƶ) = n_0 e^(-mgz//kT)` Thuis means that the cylinder contains `int _0^oo n(Ƶ) d Ƶ` =`int_0^oo n_0 e^(-mg//kT) d Ƶ = (n_0 kT)/(mg)` Particles per unit area of the base. Clearly this cannot change. Thus `n_0 kT = p_0` = pressure at the bottom of the cylinder must not change with change of temperature.

Discussion

1606.	A horizontal tube with closed ends is rotated with a constant angular velocity `omega` about a vertical axis passing through one of its ends. The tube contains carbon dioxide at a temperature `T = 300 K`. The length of the tube is `l = 100 cm`. Find the value `omega` at which the ratio of molecular concentrations at the opposite ends of the tube is equal to `eta = 2.0`.
Answer» The potential energy associated with each molecule is : `-(1)/(2) m omega^2 r^2` and there is a concentration variation `n(r) = n_0 exp((m omega^2 r^2)/(2 kT)) = n_0 exp ((m omega^2 r^2)/(2 RT))` Thus `eta = exp ((M omega^2 l^2)/(2 R T)`or `omega = sqrt((2 R T)/(Ml^2) 1n eta)` Using `M = 12 + 32 = 44 gm, l = 100 cm, R = 8.31 xx 10^7 (erg)/(.^@K), T = 300`, we get `omega = 280` radians per second.

Discussion

1607.	A tall vertical vessel contains a gas composed of two kinds of molecules of masses `m_1` and `m_2` . With `m_2 gt m_1`. The concentrations of these molecules at the bottom of the vessel are equal to `n_1` and `n_2` respectively, with `n_2 gt n_1`. Assuming the temperature `T` and the free-fall acceleration `g` to be independent of the height, find the height at which the concentrations of these kinds of molecules are equal.
Answer» `n_1(h) = n_1 e^(-m_1 gh//kt), n_2 (h) = n_2 e^(gh(m_1 - m_2)//kT)` They are equal at a height `h` where `(n_1)/(n_2) = e^(gh(m_1 - m_2)//kT)` or `h = (kT)/(g) (1 n n_1 - 1n n_2)/(m_1 - m_2)`.

Discussion

1608.	Give the value of resonance energy for using these data :A. 1 kcal`//`moleB. 2 kcal`//`moleC. 3 kcal`//`moleD. 4 kcal`//`mole
Answer» Correct Answer - B

Discussion

1609.	A reaction at 400 K with approximate `DeltaG^(@)=3207 J//"mole"`, consists of 4 mole of substance A, 2 mole of substance B and 3 mole of substance C at equilibrium in 1 litre container, then which of the following is correctly balanced reaction ? [Take : log 3 =0.48, log 2= 0.3 in x =2.30 log x R=8.3 `J//"mole"//K`]A. `2A(g)hArrB(g)+C(g)`B. `A(g)hArrB(g)+C(g)`C. `A(g)+B(g)hArr2C(g)`D. `A(g)+C(g)hArr2B(g)`
Answer» Correct Answer - A

Discussion

1610.	What is the free energy for the process `H_(2)O(l,-10^(@)C, 0.28Pa,1"mole")toH_(2)O` `(s,-10^(@)C,0.26Pa, 1"mole")` Given that vapor pressure of water and ice at `-10^(@)C` is 0.28and0.26Pa respectivelyA. `Rxx263xxIn(14)/(13)`B. `Rxx263xxIn(13)/(14)`C. `-Rxx10xxIn(13)/(14)`D. zero
Answer» Correct Answer - B

Discussion

1611.	For a reaction: `2A(g)+4B(g)to 5C(g)+2D(l)DeltaE^(@)=40kcal//"mole" and DeltaS^(@)=+200cal//K"mole"` The value of `DeltaG_(200)^(@)` will be.
Answer» Correct Answer - C

Discussion

1612.	Which of the following parameters is correct reagarding adsorption of gases over solid?A. `deltaS_(system)gt0`B. `deltaS_("surrounding")gt0`C. `deltaG gt0`D. `deltaH gt0`
Answer» Correct Answer - B

Discussion

1613.	Identify the option which is correct :A. for an adiabatic process, entropy, of system must remains constant is always.B. for porcess to be spontaneous Gibbs free energy should decreses.C. for any substance heat required for melting will always be more than heat required for vaporisation since solids will have stronger interaction.D. `DeltaS_(f)^(@)of H_(2)O(l)is lt0.`
Answer» Correct Answer - D

Discussion

1614.	Which one of the following statement is false ?A. For a fixed amount of substance, entropy of a substance in agaseous phaseB. spontanous process always occur very rapidlyC. for a fixed amount of a solid ,`DeltaS` sublimation of a solid may balways be positiveD. If `DeltaGlt0` then it is not necessary that process will be spontaneous always
Answer» Correct Answer - B

Discussion

1615.	The change in entropy of the system moles of a diatomic ideal gas is heated from 400 K to 800 K under constant pressure:A. 3 R in 2B. 7 R in 2C. 5 R in 2D. R in 2
Answer» Correct Answer - B

Discussion

1616.	For a reaction: `2A(g)+B(g)toC(g), DeltaU^(@)=30kcal//"mole" , DeltaS^(@)=100cal//K"mole". "What will be the value of" DeltaG_(reaction)^(@)at 400K?` `[R=2cal//K"mole"]" "[DeltaC_(p_(reaction))=0]`
Answer» Correct Answer - C

Discussion

1617.	For the reaction : `2A(g)+B(g)toC(g)` Change in enthalpy is 30` "kcal"//"mole whereas" DeltaU " is " 32 "kcal"//"mole"`at certain temperature. Calculate the work done (in kcal) when 4 mole of A reacts with excess of B at constant pressure and same temperature.
Answer» Correct Answer - 4

Discussion

1618.	Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally Insulated. The stopcock is now opened. Answer the following. 1. What is the final pressure of the gas in A and B? 2. What is the change in Internal energy of the gas? 3. What is the change in temperature of the gas? 4. Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?
Answer» 1. The above process is a case of free expansion of gas. As soon as the stopcock is removed the gas expands to a total volume of twice its original value (as V_A = V_B). From Boyle’s law, we have Pα₁/v Since V doubles after the stopcock is removed. P reduces to one half the original value. At STP, pressure of the gas is 100 kPa = 1 Bar ∴ Pressure after expansion = 50 kPa = 0.5 Bar. 2. From the first law of thermodynamics ∆ Q = ∆ U + ∆ W, Since the process does not involve any work done by the gas such as moving a piston, and no heat is exchanged, ∆ Q = ∆ W = 0 ∴ ∆ U = O ∴ There is no change in internal energy of the gas. 3. Since the internal energy of the gas is fixed for the given process, the temperature of the gas also does not change. ie, ∆ T = ∆ U and if ∆ U = 0, then ∆ T = 0 4. In case of free expansion of gas, the gas does not go through states of thermodynamic equilibrium before reaching the final state. There fore, the thermodynamic parameters such as pressure, volume and temperature are not well defined for these intermediate states and hence, they do not lie on the P-V-T surface for the gas.

1618.

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally Insulated. The stopcock is now opened. Answer the following. 1. What is the final pressure of the gas in A and B? 2. What is the change in Internal energy of the gas? 3. What is the change in temperature of the gas? 4. Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?

Answer»

1. The above process is a case of free expansion of gas. As soon as the stopcock is removed the gas expands to a total volume of twice its original value (as V_A = V_B).

From Boyle’s law, we have Pα₁/v

Since V doubles after the stopcock is removed. P reduces to one half the original value.

At STP, pressure of the gas is 100 kPa = 1 Bar

∴ Pressure after expansion = 50 kPa = 0.5 Bar.

2. From the first law of thermodynamics

∆ Q = ∆ U + ∆ W,

Since the process does not involve any work done by the gas such as moving a piston, and no heat is exchanged,

∆ Q = ∆ W = 0

∴ ∆ U = O

∴ There is no change in internal energy of the gas.

3. Since the internal energy of the gas is fixed for the given process, the temperature of the gas also does not change.

ie, ∆ T = ∆ U and if ∆ U = 0, then ∆ T = 0

4. In case of free expansion of gas, the gas does not go through states of thermodynamic equilibrium before reaching the final state. There fore, the thermodynamic parameters such as pressure, volume and temperature are not well defined for these intermediate states and hence, they do not lie on the P-V-T surface for the gas.

Discussion

1619.	For the reaction `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g)` Which one of the statement is correct at constant T and P ?A. `Delta H = Delta E`B. `Delta H lt Delta E`C. `Delta H gt Delta E`D. `Delta H` is independent of physical state of reactants
Answer» Correct Answer - B

Discussion

1620.	For a reaction `2X(s)+2Y(s)rarr 2Cl(l)+D(g)` The `q_(p)` at `27^(@)C` is -28 K Cal. `mol^(-1)`. The `q_(V)` is ___________ K. Cal. `mol^(-1)` :-A. `-27.4`B. `+27.4`C. `-28.6`D. `28.6`
Answer» Correct Answer - C

Discussion

1621.	Heat of reaction for, `CO(g)+1//2O_(2)(g)rarr CO_(2)(g)` at constant V is -67.71 K cal at `17^(@)C` . The heat of reaction at constant P at `17^(@)C` is :-A. `-68.0` K calB. `+68.0` K calC. `-67.4` K calD. None
Answer» Correct Answer - A

Discussion

1622.	For the system `S (s) + O_(2)(g) rarr SO_(2)(g)` ?A. `Delta H = Delta E`B. `Delta H gt Delta E`C. `Delta E gt Delta H`D. `Delta H = 0`
Answer» Correct Answer - A

Discussion

1623.	The difference in `Delta H` and `Delta E` for the combustion of methane at `25^(@)C` would be :-A. ZeroB. `2xx298xx -2` calsC. `2xx298xx -3` calsD. `2xx25xx -3` cals
Answer» Correct Answer - B

Discussion

1624.	Which of the following reactions is `Delta H` less than `Delta E` ?A. `C_(12)H_(22)O_(11)(s)+6O_(2)(g)rarr 6CO_(2)(g)+6H_(2)O(l)`B. `2SO_(2)(g)+O_(2)(g)rarr 2SO_(3)(g)`C. `N_(2)O_(4)(g)rarr 2NO_(2)(g)`D. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`
Answer» Correct Answer - B

Discussion

1625.	For which change `Delta H ne Delta E` :-A. `H_(2)(g)+I_(2)(g)hArr 2HI(g)`B. `HCl(l)+NaOH(l)rarr NaCl(s)+H_(2)O(l)`C. `C(s)+O_(2)(g)rarr CO_(2)(g)`D. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`
Answer» Correct Answer - D

Discussion

1626.	The heat of neturalization is maximum whenA. ammonium hydroxide is neutralized by acetic acidB. ammonium hydroxide is neutralized by hydrochloric acidC. sodium hydroxide is neutralized by formic acidD. sodium hydroxide is neutrailized by hydrochloric acid
Answer» Enthalpy of neutralization is maximum whenever any strong acid reacts with any strong base as both are completely ionized. In case either acid or base or both are weak, the net amount of energy released is less as certain amount of energy is used up ionize the weak electroyle.

Discussion

1627.	Consider a gas enclosed in a container. If two divide the container into three equal parts partition, then which of each of the following properties of gas will have the same value is each of the compartment?A. Internal energyB. EnthalpyC. volumeD. Temperature
Answer» Correct Answer - D Temperature is an intensive proterty, thus, its magnitude is independent of the quantitly of size of matter. Internal energy, enthaply and volume are extensive properties and, hence, their magnitudes depends on the quantity or size of matter.

Discussion

1628.	Heat exchanged in a chemical in a chamical reaction at constant temperature and pressure is calledA. entropyB. EnthalpyC. internal energyD. free energy
Answer» Correct Answer - B According to thermodynamics, `Delta H = Delta U + P Delta V` (`T` and `P` constant) `= q + ( - P Delta V) + P Delta V` `= q` (`T`, `P` constant) Thus, heat exchanged by the system with its surroundings at constant `T` and `P` is called enthalpy change. Hence, `Delta H` is sometimes called the heat of reaction at constant `P`.

Discussion

1629.	The process of evaporation of a liquid is accompanied byA. Increase in ethalpyB. Decrease in free energyC. Increase in entropyD. All
Answer» Correct Answer - D

Discussion

1630.	When the value of entropy is greater then the ability for useful work is :A. MaximumB. MinimumC. MediumD. None of these
Answer» Correct Answer - B

Discussion

1631.	If `Delta G^(@)gt 0` for a reaction then :A. `K_(P) gt 1`B. `K_(P) lt 1`C. The products predominate in the equilibrium mixtureD. None
Answer» Correct Answer - B

Discussion

1632.	A reaction `A +B rarr C+D +q` is found to have a positive entropy change, the reaction will be:A. Possible at high temperatureB. Possible only at low temperatureC. Not possible at any temperatureD. Possible at any temperature
Answer» Correct Answer - D

Discussion

1633.	The Vant Hoff equation is :A. `Delta G^(@)=RT log_(e )K_(P)`B. `-Delta G^(@)=RT log_(e )K_(P)`C. `Delta G^(@)=RT^(2)lnK_(P)`D. None
Answer» Correct Answer - B

Discussion

1634.	Equilibrium constant of a reaction is elated to :A. Standard free energy change `Delta G^(@)`B. Free energy change `Delta G`C. Entropy changeD. None
Answer» Correct Answer - A

Discussion

1635.	Calculate the standard entropy change for the following reaction (∆Hf°), given the standard entropies of COf (g), C(s), O2(g) as 213.6, 5.740 and 205 JK-1 respectively.
Answer» C_(g) + O₂_(g) → CO₂_(g) ∆S_r° = ∑S_products° – S_reactants° ∆S_r° = {\(S{_{CO}{_2}}\)°} – {S_c° + \(S{_O{_2}}\)°} ∆S_r° = 213.6 – [5.74 + 205] ∆S_r° = 213.6 – [210.74] ∆S_r° = 2.86 JK^-1.

1635.

Calculate the standard entropy change for the following reaction (∆Hf°), given the standard entropies of COf (g), C(s), O2(g) as 213.6, 5.740 and 205 JK-1 respectively.

Answer»

C_(g) + O₂_(g) → CO₂_(g)

∆S_r° = ∑S_products° – S_reactants°

∆S_r° = {\(S{_{CO}{_2}}\)°} – {S_c° + \(S{_O{_2}}\)°}

∆S_r° = 213.6 – [5.74 + 205]

∆S_r° = 213.6 – [210.74]

∆S_r° = 2.86 JK^-1.

Discussion

1636.	For the process `H_(2)O(l)toH_(2)O(g)` at `t=100^(@)C` and 1 atmosphere pressure, the correct choice is:A. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") lt 0`
Answer» Correct Answer - B `H_(2)O(l) rarr H_(2)O(g) "at" T=100 ^(@) C, 1atm` equilibrium exists . `therefore DeltaG=0 , DeltaH-TDeltaS=0` `DeltaH=TDeltaS gt 0` for system m since evaportation is endothermic `(DeltaS)_("system") gt 0 , also (DeltaS)_("surrounding")=(q_("surr"))/(T_("surr"))` Heat gained by system = heat lost by surroundings `therefore q_("surr") lt 0 therefore (DeltaS)_("surr") lt 0`

Discussion

1637.	Match the thermodynamic processes given under column I with the expressions given under column II.
Answer» Correct Answer - A::B::C::D `(A) H_(2)O(l) rarr H_(2)(s)` at 273K. & 1 atm `DeltaH =-ve =q` `DeltaS_("sys") lt 0, DeltaG=0.` `w ne 0` (as water expands on freezing ), `DeltaU ne 0` (B) Free expansion of ideal gas. `" "q=0` `w=0` `DeltaU=0` `DeltaS_("sys") gt 0` `DeltaG lt 0` (C) Mixing of equal volumn of ideal gases at constant pressure & temp in an isolated container `q=0, w=0, DeltaU=0, DeltaS_("sys") gt 0, DeltaG lt 0` `H_(2)(g)300K underset("Heating 1 atm") overset("Revesible") rarr 600 K underset("Cooling 1 atm") overset("Reversible")rarr 300 K .` `q=0, w=0, DeltaU=0 , Delta=0 DeltaS_("sys") =0`

Discussion

1638.	The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at `25^(@)C` are `-156` and `+49 kJ mol^(-1)`, respectively. The standard enthaly of hydrogenation of cyclohenxene (l) at `25^(@)C` is `-119 kJ mol^(-1)` Use this data to estimate the magnitude of the resonance enegry of benzene.A. `152 kJ mol^(-1)`B. `-240 kJ mol^(-1)`C. `-152 kJ mol^(-1)`D. `240 kJ mol^(-1)`
Answer» Correct Answer - C `6C + 6H_(2) rarr C_(6) H_(12) , -156` ....(1) `6C + 3H_(2) rarr C_(6)H_(6) , + 49`...(2) `C_(6) H_(10) + H_(2) rarr C_(6)H_(12) , -119` `rArr C_(6)H_(6) + 3H_(2) rarr C_(6) H_(12) , -119 xx 3`...(3) ltbr. [Obseved ]from (1) and (2) `C_(6)H_(6) + 3H_(2) rarr C_(6) H_(12) -205` [Calculated] `,.` Reasonance energy `= (Delta H)_("obs") ~ (Delta H)_("cal")`

Discussion

1639.	Oxidizing power of chlorine in aqueous solution can be determined by the parameters indicated below`:` `(1)/(2)Cl_(2)overset((1)/(2) Delta_("disso")H^(c-))(rarr)Cl(g) overset(Delta_(cg) H^(c-))(rarr) Cl^(-)(g)overset(Delta_(hyd)H^(c-))(rarr) Cl^(-) (aq)` The energy involved in the conversion of `(1)/(2)Cl_(2)(g)` to`Cl^(-) (aq)` ( using the data, `Delta_(disso) H_(Cl_(2))^(c-) = 240 kJmol^(-1),Delta_(eg)H = - 349 kJ mol^(-1) Delta_(hyd)H_(cl) ^(c-) = - 381kJ mol^(-1))` will beA. `- 850kJ mol^(-1)`B. `+120kJ mol^(-1)`C. ` +152kJ mol^(-1)`D. `- 610 kJmol^(-1)`
Answer» Correct Answer - D For the process `(1)/(2) Cl_(2)(g) rarr Cl^(-) (aq) `,using the given steps, `DeltaH =(1)/(2)Delta_(diss)H_(Cl_(2))+ Delta_(eg) Cl_ +Delta_(hyd) H_(Cl^(-))` `= ( 240)/(2) - 349-381kJmol^(-1)` `=120 -730=-610 kJ mol^(-1)`

Discussion

1640.	For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct optionA. Work done at 600 K is 20 times the work done at 300 KB. work done at 300K is twice the work done at 600 KC. work done at 600 K is twice the work done at 300 KD. `Delta U = 0` in both cases
Answer» Correct Answer - C::D `w = -nRT ln.(V_(f))/(V_(i))` Here, n, R and `ln.(V_(f))/(V_(i))` are same in both the cases therefore `(w_(2))/(w_(1)) = (T_(2))/(T_(1))` or `(w(600K))/(w(300K)) = (600K)/(300K) = 2`

Discussion

1641.	For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression `w= - nRT ln . (V_(f))/( V_(i)0` . A sample containig 1.0 molof an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments . The expansions is carried out at 300 K and at600 K respectively. Choose the correct option.A. Work done at 600 K is 20 times at the work done at 300KB. Work done at 300 K is twice the work done at 600 KC. Work done at 600 K is twice the work done at 300 KD. `DeltaU=0`in both cases
Answer» Correct Answer - c,d `w= -nRT ln. (V_(1))/( V_(2)). `The factors n , R and `ln. (V_(f))/( V_(i))` are same in both the cases. Hence, `(w_(2))/(w_(1)) = (T_(2))/(T_(1)), i.e., (.^(w)600K)/(.^(w)300K)= ( 600K)/( 300K) =2 , i.e., ` work done at 600K is twice the work done at 300K. As each case involves isothermal expansion of an ideal gas, there is no change in internal energy, i.e., `DeltaU =0`

Discussion

1642.	For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct optionA. work done at 600 K is 20 times the work done at 300 KB. work done at 300 K is twice the work done at 600 KC. work done at 600 K is twice the work done at 300 KD. `Delta U = 0` in both cases.
Answer» Correct Answer - C::D Given that, the work of reversible expansion under isothernal condition can be calculated by using the expression `W = - nRT ln. (V_(1))/(V_(1))` `V_(1) = 10 V_(i)` `T_(2) = 600 K` `T_(1) = 300 K` Putting these values in above expression `W_(600 K) = 1 xx R xx 600 ln. (10)/(1)` `W_(300 K) = 1 xx R xx 300 K ln. (10)/(1)` Ration `= (W_(600 K))/(W_(300 K)) = (1 xx R xx 600 K ln.(10)/(1))/(1 xx R xx 300 K ln. (10)/(1)) = (600)/(300) = 2` For isothermal expansion of ideal gases, `Delta U = 0`. Since, temperature is contant this means there is no change in internal energy. Therefore, `Delta U = 0`

Discussion

1643.	Hydrolysis of sucrose gives Sucrose `+ H_(2)O hArr `Glucose `+` Fructose Equilibrium constant `K_(c)` for the reaction is `2 xx 10^(13)` at 300 K. Calculate `Delta G^(@)` at 300 K.
Answer» `DeltaG^(@) = - 2.303 RT log K_(c) =- 2.303 xx ( 8.314JK^(-1) mol^(-1)) xx 300 K xx log ( 2 xx 10^(13))` `= - 2.303 xx 8.314 xx 300 xx 13.301 J mol ^(-1) = - 76402 J mol^(-1) = -764 xx 10^(4) J mol^(-1)`

Discussion

1644.	Does specific heat of a gas possess a unique value?
Answer» No. It is of two types for gases, specific heat at constant pressure c_p and specific heat at constant volume c_v.

Discussion

1645.	What is the principle of calorimetry?
Answer» Heat lost by the hot body is equal to the heat gained by the cold body, i.e., Heat lost = Heat gained.

Discussion

1646.	What is pyrometer?
Answer» Pyrometer is a radiation thermometer. It can be used to measure the temperature of a furnace or sun, etc.

Discussion

1647.	The equilibrium constant for a reaction is one or more if ∆G° for it is less than zero. Explain.
Answer» ―∆G° = RT ln K, thus if ∆G° is less than zero. i.e., it is negative, then ln K will be positive and hence K will be greater than one.

Discussion

1648.	In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as `P prop T^(C )`, Where `C` is equal to:
Answer» Correct Answer - D

Discussion

1649.	In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gasA. temperature will decreaseB. volume will increaseC. pressure will remain constantD. temperature will increase
Answer» Correct Answer - A From first law of thermodynamics, `dQ= dU+dW` `dQ=dU, as dW=0` As `dQlt0, therfore, dUlt0` `:.` temperature will decrease

Discussion

1650.	In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gasA. The temperature will decreaseB. The volume will increaseC. The pressure will remain constantD. The temperature will remain cosntant
Answer» Correct Answer - A::C

Discussion

Explore topic-wise InterviewSolutions in .

From the Maxwell distribution function frind `lt lt v_x^2 gt gt`, the mean value of the squared `v_x` projection of the molecular velocity in a gas at a temperature `T`. The mass of each molecule is equal to `m`.

Using the Maxwell distribution function, determine the pressure exterted by gas on a wall, if the gas temperature is `T` and the concentration of molecules is `n`.

A very tall vertical cylinder contains a gas at a temperature true `T`. Assuming the gravitational field to be uniform, find the mean value of the potential energy of the gas molecules. Does this value depend on whether the gas consists of one kind of molecules or of serveral kinds ?

A very tall vertical cylinder contains carbon dioxide at a certain temperature `T`. Assuming the gravitational field to be uniform, find how the gas pressure on the bottom of the vessel will change when the gas temperature increases `eta` times.

Give the value of resonance energy for using these data :A. 1 kcal`//`moleB. 2 kcal`//`moleC. 3 kcal`//`moleD. 4 kcal`//`mole

What is the free energy for the process `H_(2)O(l,-10^(@)C, 0.28Pa,1"mole")toH_(2)O` `(s,-10^(@)C,0.26Pa, 1"mole")` Given that vapor pressure of water and ice at `-10^(@)C` is 0.28and0.26Pa respectivelyA. `Rxx263xxIn(14)/(13)`B. `Rxx263xxIn(13)/(14)`C. `-Rxx10xxIn(13)/(14)`D. zero

For a reaction: `2A(g)+4B(g)to 5C(g)+2D(l)DeltaE^(@)=40kcal//"mole" and DeltaS^(@)=+200cal//K"mole"` The value of `DeltaG_(200)^(@)` will be.

Which of the following parameters is correct reagarding adsorption of gases over solid?A. `deltaS_(system)gt0`B. `deltaS_("surrounding")gt0`C. `deltaG gt0`D. `deltaH gt0`

The change in entropy of the system moles of a diatomic ideal gas is heated from 400 K to 800 K under constant pressure:A. 3 R in 2B. 7 R in 2C. 5 R in 2D. R in 2

For a reaction: `2A(g)+B(g)toC(g), DeltaU^(@)=30kcal//"mole" , DeltaS^(@)=100cal//K"mole". "What will be the value of" DeltaG_(reaction)^(@)at 400K?` `[R=2cal//K"mole"]" "[DeltaC_(p_(reaction))=0]`

For the reaction : `2A(g)+B(g)toC(g)` Change in enthalpy is 30` "kcal"//"mole whereas" DeltaU " is " 32 "kcal"//"mole"`at certain temperature. Calculate the work done (in kcal) when 4 mole of A reacts with excess of B at constant pressure and same temperature.

For the reaction `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g)` Which one of the statement is correct at constant T and P ?A. `Delta H = Delta E`B. `Delta H lt Delta E`C. `Delta H gt Delta E`D. `Delta H` is independent of physical state of reactants

For a reaction `2X(s)+2Y(s)rarr 2Cl(l)+D(g)` The `q_(p)` at `27^(@)C` is -28 K Cal. `mol^(-1)`. The `q_(V)` is ___________ K. Cal. `mol^(-1)` :-A. `-27.4`B. `+27.4`C. `-28.6`D. `28.6`

Heat of reaction for, `CO(g)+1//2O_(2)(g)rarr CO_(2)(g)` at constant V is -67.71 K cal at `17^(@)C` . The heat of reaction at constant P at `17^(@)C` is :-A. `-68.0` K calB. `+68.0` K calC. `-67.4` K calD. None

For the system `S (s) + O_(2)(g) rarr SO_(2)(g)` ?A. `Delta H = Delta E`B. `Delta H gt Delta E`C. `Delta E gt Delta H`D. `Delta H = 0`

The difference in `Delta H` and `Delta E` for the combustion of methane at `25^(@)C` would be :-A. ZeroB. `2xx298xx -2` calsC. `2xx298xx -3` calsD. `2xx25xx -3` cals

Which of the following reactions is `Delta H` less than `Delta E` ?A. `C_(12)H_(22)O_(11)(s)+6O_(2)(g)rarr 6CO_(2)(g)+6H_(2)O(l)`B. `2SO_(2)(g)+O_(2)(g)rarr 2SO_(3)(g)`C. `N_(2)O_(4)(g)rarr 2NO_(2)(g)`D. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`

For which change `Delta H ne Delta E` :-A. `H_(2)(g)+I_(2)(g)hArr 2HI(g)`B. `HCl(l)+NaOH(l)rarr NaCl(s)+H_(2)O(l)`C. `C(s)+O_(2)(g)rarr CO_(2)(g)`D. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`

The heat of neturalization is maximum whenA. ammonium hydroxide is neutralized by acetic acidB. ammonium hydroxide is neutralized by hydrochloric acidC. sodium hydroxide is neutralized by formic acidD. sodium hydroxide is neutrailized by hydrochloric acid

Consider a gas enclosed in a container. If two divide the container into three equal parts partition, then which of each of the following properties of gas will have the same value is each of the compartment?A. Internal energyB. EnthalpyC. volumeD. Temperature

Heat exchanged in a chemical in a chamical reaction at constant temperature and pressure is calledA. entropyB. EnthalpyC. internal energyD. free energy

The process of evaporation of a liquid is accompanied byA. Increase in ethalpyB. Decrease in free energyC. Increase in entropyD. All

When the value of entropy is greater then the ability for useful work is :A. MaximumB. MinimumC. MediumD. None of these

If `Delta G^(@)gt 0` for a reaction then :A. `K_(P) gt 1`B. `K_(P) lt 1`C. The products predominate in the equilibrium mixtureD. None

A reaction `A +B rarr C+D +q` is found to have a positive entropy change, the reaction will be:A. Possible at high temperatureB. Possible only at low temperatureC. Not possible at any temperatureD. Possible at any temperature

The Vant Hoff equation is :A. `Delta G^(@)=RT log_(e )K_(P)`B. `-Delta G^(@)=RT log_(e )K_(P)`C. `Delta G^(@)=RT^(2)lnK_(P)`D. None

Equilibrium constant of a reaction is elated to :A. Standard free energy change `Delta G^(@)`B. Free energy change `Delta G`C. Entropy changeD. None

Calculate the standard entropy change for the following reaction (∆Hf°), given the standard entropies of COf (g), C(s), O2(g) as 213.6, 5.740 and 205 JK-1 respectively.

Match the thermodynamic processes given under column I with the expressions given under column II.

Hydrolysis of sucrose gives Sucrose `+ H_(2)O hArr `Glucose `+` Fructose Equilibrium constant `K_(c)` for the reaction is `2 xx 10^(13)` at 300 K. Calculate `Delta G^(@)` at 300 K.

Does specific heat of a gas possess a unique value?

What is the principle of calorimetry?

What is pyrometer?

The equilibrium constant for a reaction is one or more if ∆G° for it is less than zero. Explain.

In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as `P prop T^(C )`, Where `C` is equal to:

In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gasA. temperature will decreaseB. volume will increaseC. pressure will remain constantD. temperature will increase

In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gasA. The temperature will decreaseB. The volume will increaseC. The pressure will remain constantD. The temperature will remain cosntant