Answer to Question #88521 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Two identical gaseous systems, each containing 0.06 mol of ideal gas are at 300 K and 2.0 atm pressure. The ratio of heat capacities of the gas is 1.4. One of the gases is made to expand adiabatically and the other isothermally until they are at normal pressure. Calculate the final volumes in each case.

Answer 1

The initial volume we can get from equation of state

$$P_iV_i=\nu RT$$

$$V_i=\frac{\nu RT}{P_i}$$

For isothermal process

$$P_fV_f=P_iV_i$$

So

$$V_f=\frac{P_iV_i}{P_f}=\frac{P_i}{P_f}\frac{\nu RT}{P_i}$$ $$=\frac{2\:\rm{atm}}{1\:\rm{atm}}\frac{0.06\:\rm{mol}\times 8.31\:\rm{J/(K\: mol)}\times 300\:\rm{K}}{2\:\rm{atm}\times 101325\:\rm{Pa/atm}}$$

$$=0.0015\:\rm{m^3}=1.5\:\rm{L}$$

For adiabatic process

$$P_fV_f^{\gamma}=P_iV_i^{\gamma}$$

So

$$V_f=\left(\frac{P_i}{P_f}\right)^{1/\gamma}\frac{\nu RT}{P_i}$$

$$=\left(\frac{2\:\rm{atm}}{1\:\rm{atm}}\right)^{1/1.4}\frac{0.06\:\rm{mol}\times 8.31\:\rm{J/(K\: mol)}\times 300\:\rm{K}}{2\:\rm{atm}\times 101325\:\rm{Pa/atm}}$$

$$=0.0012\:\rm{m^3}=1.2\:\rm{L}$$