Thermodynamics - Result Question 23
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24. A mass of diatomic gas $(\gamma=1.4)$ at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $27^{\circ} C$ to $927^{\circ} C$. The pressure of the gas in final state is
======= ####24. A mass of diatomic gas $(\gamma=1.4)$ at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $27^{\circ} C$ to $927^{\circ} C$. The pressure of the gas in final state is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/thermodynamics/thermodynamics—result-question-23.md (a) $28 atm$
(b) $68.7 atm$
(c) $256 atm$
(d) $8 atm$
[2011M]
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Answer:
Correct Answer: 24. (c)
Solution:
- (c) $T_1=273+27=300 K$
$T_2=273+927=1200 K$
For adiabatic process,
$P^{1-\gamma} T^{\gamma}=$ constant
$\Rightarrow P_1{ }^{1-\gamma} T_1{ }^{\gamma}=P_2{ }^{1-\gamma} T_2{ }^{\gamma}$
$\Rightarrow(\frac{P_2}{P_1})^{1-\gamma}=(\frac{T_1}{T_2})^{\gamma} \Rightarrow(\frac{P_1}{T_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}$
$\Rightarrow(\frac{P_1}{P_2})^{1-1.4}=(\frac{1200}{300})^{1.4} \Rightarrow(\frac{P_1}{P_2})^{-0.4}=(4)^{1.4}$
$ \begin{aligned} & \Rightarrow(\frac{P_2}{P_1})^{0.4}=4^{1.4} \\ & \Rightarrow P_2=P_1 4^{(\frac{1.4}{0.4})}=P_1 4^{(\frac{7}{2})} \\ & \Rightarrow P_1(2^{7})=2 \times 128=256 atm \end{aligned} $
