Saved Bookmarks
| 1. |
We transfer 1000 J as heat to a diatomic gas, allowing the gas to expand with the pressure held constant. The gas molecules each rotate around an internal axis but do not oscillate. How much of the 1000 J goes into the increase of the gas's internal energy? Of that amount, how much goes into Delta K_( "trans") (the kinetic energy of the translational motion of the molecules) and AK (the kinetic energy of their rotational motion)? |
|
Answer» Solution :(1) The transfer of energy as heat to a gas under constant pressure is related to the resulting temperature increase `Delta T` via Eq. 21-25`(Q=nC_p Delta T )`. (2) Because the gas is diatomic with molecules undergoing rotation but not oscillation, the molar specific heat is, from Fig. 21-12 and Table 20-3, `C_p ` =(7/2)R. (3) The increase AE, in the internal energy is the same as would occur with a constant volume process resulting in the same AT. THUS, from Eq, 21-24, `Delta E_( i nt ) = nC_VDelta T`From Fig. 21-12 and Table 20-3, we see that `C_v = (5/2)`R. (4) For the same n and AT, AE is greater for a diatomic gas than for a monatomic gas because additional energy is required for rotation. Increase in `E_( i nt )`: Let.s first get the temperature change AT due to the transfer of energy as heat. From Eq. 21-25, SUBSTITUTING (7/2) R for `C_p`, we have ` Delta T= (Q )/( 1/2nR ) ` We next FIND `Delta E_(i nt )` from Eq. 21-24, substituting the molar specific heat `C_v ( = 5//2R)` for a constant volume process and using the same `Delta T`. Because we are dealing with a Diatomicgas,Let .s callthsichange`DeltaE_( i n t , d ia )` equation21-24gives us ` DeltaE_( i n t , i d a )= n C_V DeltaT= n 5/2R((Q )/( 7/2nR ))=(5)/(7 )Q` ` = 0.71428 Q = 714.3J` In words, about 71% of the energy transferred to the gas goes into the internal energy. The rest goes into the work required to increase the volume of the gas, as the gas pushes the walls of its container outward. Increase in K: If we were to increase the temperature of a monatomic gas (with the same value of n) by the amount given in Eq. 21-30, the internal energy would change by a smaller amount, call it `DeltaE_(" int , dia")`because rotational motion is not involved. To CALCULATE that smaller amount, we still use Eq. 21-24 but now we substitute the value of `C_V `, for a monatomic gas-namely,`C_v =3//2 .`so ` Detla E_( " int mon ")= n3/2R Delta T ` Substituting for `Detla `T from Eq. 21-30 leads us to ` DeltaE_( " int . mon")=n(3)/(2)R ((Q )/(n 7/2 R)) = 3/7Q ` `=0.42857Q= 428.6J.` Learn: For the monatomic gas, all this energy would go into the kinetic energy of the translational motion of the atoms. The important point here is that for a diatomic gas with the same values of n and `Delta `T, the same amount of energy goes into the kinetic energy of the translational motion of the molecules. The rest of `Delta E_(" int. dia ")`. (that is, the additional 285.7 ) goes into the rotational motion of the molecules. Thus, for the diatomic gas, ` DeltaK_("trans") = 428.6J andDelta K_(r o t)= 285.7J` |
|