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 $V_i=10^{-3}{m}^3$ changes to a final state at $P_f=(1/32)\times 10^{5}Pa$ 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. 112 J
• B. 294 J
• C. 588 J
• D. 813 J

Answer 1

Answer: (C)

$P_i=10^{5}Pa,V_i=10^{-3}{m}^3$
$P_f=\frac{1}{32}\times 10^{5}Pa,V_f=8\times 10^{-3}{m}^3$

$\because$ process is $P^3{V}^5=$ const
$P{V}^{5/3}=$ const.
for adiabatic process $P{V}^{\gamma }=$ const.
$\therefore \gamma =5/2=$ monoatomic gas.
for monoatomic gas
$C_P=\frac{5R}{2},C_V=\frac{3R}{2}$
$I{I}^{nd}$ process

heat absorbed in AB process.
$Q_1=n{C}_p\Delta T=n\frac{5}{2}R\Delta T$
$=\frac{5}{2}\left(P_2{V}_2-P_1{V}_1\right)$
$=\frac{5}{2}{P}_i\left(V_f-V_i\right)$
$=\frac{5}{2}\times 10^5\times 7\times 10^{-3}$
$=\frac{35}{2}\times 10^2=1750J$
heat above in $BC$ process $=Q_2=n{C}_V\Delta T=n\frac{3}{2}R\Delta T=\frac{3}{2}{V}_f\Delta P$
$=\frac{3}{2}\times 8\times 10^{-3}\times \left(\frac{1}{32}-1\right)\times 10^5$
$=-12\times 10^2\times \frac{31}{32}$
$=-1162.5$
Total heat $=1750-1162.5=587.5\approx 588J$