Discussion Forum

1)

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $P_{i}=10^{5}$ P_a  and volume $V_{i}=10^{-3}m^{3}$ changes to final state at  $P_{f}=(\frac{1}{32})\times 10^{5} P_{a}$  and  $V_{f}=8 \times 10^{-3} m^{3}$ in an adiabatic  quasi-static  process, such  that $P^{3} V^{5}$ =  constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps : an isobaric expansion at $P_{i}$  followed by an isochoric(isovolumetric) process at  volume $V_{f}$ . The amount of heat supplied to the system  in the two step process is approximately

 

Answer:

Option C

Explanation:

In the first process $p_{i}V_{i}^{\gamma}=p_{f}V_{f}^{\gamma}$

$\Rightarrow \frac{p_{i}}{p_{f}}=(\frac{V_{f}}{V_{i}})^{\gamma}\Rightarrow32=8^{\gamma}$

$\gamma =\frac{5}{3}$ ......(i)

 For the two step process

$W= P_{i}(V_{f}-V_{i})=10^{5}(7\times 10^{-3})$

$W= 7\times 10^{2}$ J

$\triangle U= \frac{f}{2}(p_{f}V_{f}-p_{i}V_{i})$

$=\frac{I}{\gamma-1}(\frac{1}{4}\times 10^{2}-10^{2})$

$\triangle U= -\frac{3}{2}.\frac{3}{4}\times 10^{2}=-\frac{9}{8}\times 10^{2}J$

$Q-W=\triangle U$

$\Rightarrow 7\times 10^{2}-\frac{9}{8}\times 10^{2}$

$=\frac{47}{8}\times 10^{2}J =588J$