Question

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $P_i={10}^{5}Pa$ and volume $V_i={10}^{-3}Pa$ changes to a final state at $P_f=\left(\frac{1}{32}\right)\times {10}^{5}Pa$ 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

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

Answer 1

The correct option is C $588J$


In adiabatic process ideal gas equation: $P{V}^{\gamma }=Constant$
from given data, $P^3{V}^5=constant$
So, $P{V}^{\frac{5}{3}}=k$
where, $\gamma =\frac{5}{3}$
Total work done during process 1 and 2,
$W_a=\frac{P_f{V}_f-P_i{V}_i}{1-\gamma }\to W_a=\frac{132\times {10}^{-5}\times 8\times {10}^{-3}-{10}^5\times 10-3}{1-5/3}{W}_a=112.5J$

In adiabatic process the internal energy of system $\Delta Q=0so,\Delta U=-W\Rightarrow -112.5J$

Total heat supply between 1 and 2
$q_1+q_2=\Delta U+P_i(V_f-V_i)=-112.5+{10}^5(8-1)\times {10}^{-3}=587.5J$