Answer to Question #208512 in Molecular Physics | Thermodynamics for sbu

Question

An ideal gas with gamma=1.30 occupies 7.0 L at 300 K and 200 kPa pressure.

It is compressed adiabatically to 1/7 of its original volume, then cooled at constant volume to 300 K, and finally allowed to expand isothermally to 7.0 L.

Answer 1

This process is divided into three steps between the states that the gas is reaching along the process. First, we can calculate the number of mole of gas using the ideal gas law and the conditions at the start:

$n=\frac{pV}{RT}=\frac{(200000\,Pa)(0.007\,m^3)}{(8.314\,J/molK)(300\,K)}$ $=0.5613$mol of gas

Then, we calculate the work (as $W=\int pdVW=∫pdV$) on all three steps the gas goes through. For the first step (the adiabatic compression) the work will be equal to the change in the internal heat, and we also calculate the new molar heat capacity at constant volume with the gamma value:

$W_{12} = \Delta U = \int nC_Vd T = nC_V\Delta T=n C_V(T_2-T_1)$

for this gas we have $C_V = \frac{R}{\gamma-1}=\frac{8.314\,J/mol\cdot K}{1.30-1.00}=27.713\,\frac{J}{mol\cdot K}$

and we will also have to use the equations that relate volume and temperature to find T2, where

$\frac{T_2}{T_1}=(\frac{V_1}{V_2})^{\gamma-1} \implies T_2=T_1(\frac{V_1}{V_2})^{\gamma-1}=(300 K)(\frac{7\,\cancel{L}}{1\,\cancel{L}})^{1.30-1.00}$

$T_2=537.84\,K\,\implies$

with that information we proceed to find W12:

$W_{12} =(0.5613\,\cancel{mol})(27.713\,\frac{J}{\cancel{mol\cdot K}})(537.84-300)\cancel{K}$

$W_{12}=3699.674\,J$

For the second step (2-3), since the process goes under constant volume when the gas is cooled down, then W23 = 0 J (this because dV=0).

The third step goes under the same temperature that was reached before (300 K) and since we have the start volume (1L) and final volume (7 L) of the steps 3-1 we can calculate the work done:

$W_{31}=\int pdV=\int nRT\frac{dV}{V}=nRT\ln(\frac{V_f}{V_i})$

$W_{31} =(0.5613\,\cancel{mol})(8.314\,\frac{J}{\cancel{mol\cdot K}})(300\,\cancel{K})\ln(\frac{7\,\cancel{L}}{1\,\cancel{L}})$

$W_{31} =2724.263\,J$

Then, the total work done by the gas is the sum of all three processes:

$W_T=W_{12}+W_{23}+W_{31}=(3699.674+0+2724.263)JW_T=6423.937\,J =6423.937\ J$

Reference:

Castellan, G. W. (1983). Physical Chemistry. Ed.