Heat And Thermodynamics Ques 133
- 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 $V _1=10^{-3} m^{3}$ 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
(2016 Adv.)
(a) $112 J$
(b) $294 J$
(c) $588 J$
(d) $813 J$
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Answer:
Correct Answer: 133.(c)
Solution:
Formula:
- In the first process : $p _i V _i^{\gamma}=p _f V _f^{\gamma}$
$$ \begin{aligned} \Rightarrow \quad \frac{p _i}{p _f} & =\left(\frac{V _f}{V _i}\right)^{\gamma} \Rightarrow 32=8^{\gamma} \\ \gamma & =\frac{5}{3} \cdots(i) \end{aligned} $$
For the two step process
$$ \begin{aligned} W & =p _i\left(V _f-V _i\right)=10^{5}\left(7 \times 10^{-3}\right)=7 \times 10^{2} J \\ \Delta U & =\frac{f}{2}\left(p _f V _f-p _i V _i\right)=\frac{1}{\gamma-1}\left(\frac{1}{4} \times 10^{2}-10^{2}\right) \\ \Delta U & =-\frac{3}{2} \cdot \frac{3}{4} \times 10^{2}=-\frac{9}{8} \times 10^{2} J \\ Q-W & =\Delta U \\ \Rightarrow \quad Q & =7 \times 10^{2}-\frac{9}{8} \times 10^{2}=\frac{47}{8} \times 10^{2} J=588 J \end{aligned} $$
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