Question

A gas is enclosed in a cylinder with a movable friction less piston. Its initial thermodynamic state at pressure $P_i={10}^{5}Pa$ and volume $V_i={10}^{-3}{m}^3$ changes to a final state as $P_j=\left(1/32\right){\times 10}^{5}pa$ and $V_j=8\times {10}^{-3}{m}^3$ 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 $V_j$. The amount of heat supplied to the system in the two-step process is approximately.

• A. $112J$
• B. $294J$
• C. $588J$
• D. $813J$

Answer 1

The correct option is D $813J$

Work done in adiabatic process,

$\Rightarrow W=\frac{P_1{V}_1-P_2{V}_2}{r-1}$

$=\frac{{10}^5{10}^{-3}-\frac{1}{3^2}\times {10}^5\times 8\times {10}^{-3}}{\frac{5}{3}-1}$

$=112.5J$

For isobaric process,

$\Rightarrow W=P(V_2-V_1)$

$={10}^5(8\times {10}^{-3}-1\times {10}^{-3})$

$=700J$

For isochoric process -

Since volume is constant no work is done.

$\Rightarrow w=0$

Total work done :

$=112.5+700+0$

$=812.5J$

$\approx 813J$

Hence, the answer is $813J.$