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The pressure $\mathrm{P}_{1}$ and density $\mathrm{d}_{1}$ of diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes suddenly to $\mathrm{P}_{2}\left(>\mathrm{P}_{1}\right)$ and $\mathrm{d}_{2}$ respectively during an adiabatic process. The temperature of the gas increases and becomes__________ times of its initial temperature.

(given $\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=32$ )
Option: 1
4

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

For adiabatic process

$\mathrm{PV^{v}=P^{1-v}T^{v}=k}$

Also for a given mass of gas

$\mathrm{v\: \alpha \: \frac{1}{d}}$

$\mathrm{d \rightarrow \text { density of gas }}$

$\mathrm{p \times \frac{1}{d^{v}}=k}$

$\mathrm{p \propto d^v}$

$\mathrm{\therefore {d^{v-v^2} T^{v}}=K}$

$\mathrm{d_1^{v-v^2} T_1^v=d_2^{v-v^2 } T_2^v}$

$\mathrm{\frac{d_2^{v^2-v}}{d_1^{v^2-v}}=\left(\frac{T_2}{T_1}\right)^{v}}$

$\mathrm{\left ( \frac{d_{2}}{d_{1}} \right )^{v-1}=\left ( \frac{T_{2}}{T_{1}} \right )}$

$\mathrm{\left ( 32 \right )^{2/5}=\frac{T_{2}}{T_{1}}}$

$\mathrm{\frac{T_{2}}{T_{1}}=4}$

The temperature of the gas increases and becomes 4 times of its initial temperature

Posted by

Ajit Kumar Dubey

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