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
$PV ^{5 / 3}=$ const.
for adiabatic process $PV ^{\gamma}=$ const.
$\therefore \gamma=5 / 2=$ monoatomic gas.
for monoatomic gas
$C _{ P }=\frac{5 R }{2}, C _{ V }=\frac{3 R }{2}$
$II ^{nd}$ process

heat absorbed in AB process.
$Q _{1}= nC _{ p } \Delta T = n \frac{5}{2} R \Delta T$
$=\frac{5}{2}\left( P _{2} V _{2}- P _{1} V _{1}\right)$
$=\frac{5}{2} P _{ i }\left( V _{ f }- V _{ i }\right)$
$=\frac{5}{2} \times 10^{5} \times 7 \times 10^{-3}$
$=\frac{35}{2} \times 10^{2}=1750\, J$
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 $=1750-1162.5=587.5 \approx 588\, J$