V1 / T1 =  P2   V2 / T2   - Ideal gas law      V2 =   P1  V1  T2 / (T1  P2)              =  380 mm * 1 Litre * 288 K / [ 303 K  * 750 mm ]2.    P V = n R T          ideal gas law      n = R T / (P  V )  =  8.314 * (273+550) / [ 0.2 * 1.013 *10^5  * 0.350)         n =  m / M        =>          M =  m / n  =   0.05 / n 3.    P V = n R T = m R T / M       M =  m R  T / [ P V ]  = d  R T / P,    d = density            = 1.5 * 10^-3 * 8.314 * (273+65) / [750 * 1.013 * 10^5 / 760 ]            = 1.5 * 10^-3 * 8.314 * (273+65) * 760 / [750 * 1.013 * 10^5  ]4.  which conditions ?  where are they?5.open vessel   pressure P and V volume of gas remain constant.T1 = 300 K.     P1=P ,      V1 = V,     n1 = MOLES present               P  V = n1 R T1P    and V2 = V,        n2   <  n1          T2 > T1            P  V =  n2  R  T2,          n2 moles of gas remains after heating         n2 T2 = n1 T1          n2 =  (1-3/5) n1 = 2/5 * n1           =>   T2 / T1 = n1/n2 = 5/2 = 2.5          T2 = 300K * 2.5 = 750 K  ,   at this temperature only 2/5th of initial gas is present.a)  If the vessel now is heated to  T3 = 900 K, then         n3 / n1 = T1 / T3 = 1/3           =>  only 1/3 is present.  and  2/3 of air escapes out.b)  let       n4 /n1 = 1/2        n4/n1 = T1/ T4     =>   T4 = 600K====================6.V = 10L =  0.01 m^2        T = 300Kp  = m R T / (MV)p_He = 0.4  8.314 * 300 / (4 * 0.01 )     Pap_O2 = 1.6 * 8.314 * 300 / (32 * 0.01)   Pap_N2  = 1.4 * 8.314 * 300 / (28 * 0.01)  Paadd the three. to get total pressure.P = 8.314 * 300 [ 0.4/4 + 1.6 /32 + 1.4/28 ] / 0.01  ====================7.   one of N2 gas =  N = avogadro num = 6.023 * 10^23 number of molecules. volume of each MOLECULE =  4/3   Pi  r^3,        r = 2 * 10^-10 mvolume of gas at NTP,  22.4 litres = 0.0224    m^3empty space =      0.0224  -  6.023 * 10^23 * 4/3  pi   2^3   10^-30  8.P_A = 2 P_B              V_A = 2 V_B               T_A = 2 T_BP V / RT = n   =>      n_A  / n_B  =   2      =>   m_A = 2  m_B===========9.m1 = 5.40 gm        T1 = 300 K              V1 = V           P1 = Pm2 = 0.14 gm     M2 = 2        n2 = 0.07            T2 = 290 K          V2 = V         P2 = Pn1 T1  =  n2  T2     =>    n1 = 0.07 * 290 / 300     M =   m1 / n1=======================10.          P ,    d = m / V1         T1 = 300 K            P =  d R T / M        =>         d1 T1 = d2 T2                          0.75 d / d =  300 / T2          => 400 K======11V = 1 litre       P = 1 atm     T = 373 Knum of moles in STEAM:      n2 = P V / R T  = 3.381 * 10^-6  mass of this 1 litre of vapour  at 373 K,   m2 = density * Volume        =>  m2 = 0.0006 g/cc * 1000 cc =  0.6 gmsIn liquid FORM,   the water molecules weighing  0.6 gms  occupy  0.6 cc.If we assume that in water , the molecules are very CLOSELY packed (as compared to the gas vapor), then    the volume occupied by water molecules is 0.6 cc   in one litre (1000 cc) of the water vapour  at 1 atm at 373 K.