The pressure $P_{1}$ and density $d_{1}$ 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 $\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=32$ )

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Heat & Thermodynamics

The pressure

P_{1}

and density

d_{1}

of diatomic gas

(γ = \frac{7}{5})

changes suddenly to

P_{2} (> P_{1})

and

d_{2}

respectively during an adiabatic process. The temperature of the gas increases and becomes
____ times of its initial temperature.
(given

\frac{d _{2}}{d _{1}} = 32

)

Answer: 4

Solution

For an adiabatic process,

P V^{γ} = const

. Since

d = \frac{m}{V}

, we have

\frac{P}{d ^{γ}} = const

. Thus

\frac{P _{1}}{P _{2}} = (\frac{d _{1}}{d _{2}})^{γ} = (\frac{1}{32})^{7/5} = \frac{1}{128}

. Using the ideal gas law:

\frac{T _{1}}{T _{2}} = \frac{P _{1} V _{1}}{P _{2} V _{2}} = \frac{P _{1}}{P _{2}} \cdot \frac{d _{2}}{d _{1}} = \frac{1}{128} \times 32 = \frac{1}{4}

. So

T_{2} = 4 T_{1}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer

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