Heat and Thermodynamics 5 Question 57
58. An ideal gas having initial pressure $p$, volume $V$ and temperature $T$ is allowed to expand adiabatically until its volume becomes $5.66 V$ while its temperature falls to $T / 2$.
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Answer:
Correct Answer: 58. (a) f = 5 (b) W = 1.23 p V
Solution:
- (a) In adiabatic process $T V^{\gamma-1}=$ constant
$$ \begin{aligned} & \therefore \quad T V^{\gamma-1}=\left(\frac{T}{2}\right)(5.66 V)^{\gamma-1} \\ & \text { or } \quad(5.66)^{\gamma-1}=2 \\ & \therefore \quad(\gamma-1) \ln (5.66)=\ln (2) \\ & \therefore \quad \gamma-1=0.4 \\ & \text { or } \quad \gamma=1.4 \end{aligned} $$
i.e. degree of freedom, $f=5$ as $\gamma=1+\frac{2}{f}$
(b) Using, $p V^{\gamma}=$ constant
$$ \begin{aligned} p V^{1.4} & =p _f(5.66 V)^{1.4} \\ \therefore \quad p _f & =0.09 p \end{aligned} $$
Now, work done in adiabatic process
$$ \begin{aligned} W & =\frac{p _i V _i-p _f V _f}{\gamma-1} \\ & =\frac{(p V)-(0.09 p)(5.66 V)}{1.4-1} \\ & =1.23 p V \end{aligned} $$
