Thermodynamics Ques 28
28. A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be
[1996]
(a) $18^{\circ} C$
(b) $668.4^{\circ} K$
(c) $395.4^{\circ} C$
(d) $144^{\circ} C$
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Answer:
Correct Answer: 28.(b)
Solution:
- (b) Initial temperature $(T_1)=18^{\circ} C=291 K$
Let Initial volume $(V_1)=V$
Final volume $(V_2)=\frac{V}{8}$
According to adiabatic process, $T V^{-1}=$ constant
According to question, $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
$\Rightarrow T_2=293(\frac{V_1}{V_2})^{\gamma-1}$
$\Rightarrow T_2=293(8)^{\frac{7}{5}-1}=293 \times 2.297=668.4 $ $ K$
$ [\text{ For diatomic gas } \gamma=\frac{C_p}{C_v}=\frac{7}{5}] $
BETA
