Thermodynamics - Result Question 31

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33. A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be

======= ####33. A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/thermodynamics/thermodynamics—result-question-31.md (a) $18^{\circ} C$

(c) $395.4^{\circ} C$

(d) $144^{\circ} C$

(b) $668.4^{\circ} K[1996]$

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Answer:

Correct Answer: 33. (b)

Solution:

  1. (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}] $