Air is filled inside a jar which has a pressure gauge connected to it. The temperature of the air inside the jar is same as outside temperature (= T₀) but pressure (P₁) is slightly larger than the atmospheric pressure (P₀). The stopcock is quickly opened and quickly closed, so that the pressure inside the jar becomes equal to the atmospheric pressure P0. The jar is now allowed to slowly warm up to its original temperature T₀. At this time the pressure of the air inside is P₂ (P₀ < P₂ < P₁). Assume air to be an ideal gas. Calculate the ratio of specific heats (= g ) for the air, in terms of P₀, P₁ and P₂.

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$\frac{\ln (\frac{P_{1}}{P_{0}})}{\ln (\frac{P_{2}}{P_{1}})}$

$\frac{\ln (\frac{P_{0}}{P_{1}})}{\ln (\frac{P_{2}}{P_{1}})}$

$\frac{\ln (\frac{P_{0}}{P_{1}})}{\ln (\frac{P_{0}}{P_{2}})}$

$\frac{\ln (\frac{P_{0}}{P_{1}})}{\ln (\frac{P_{2}}{P_{0}})}$

answer is B.

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Detailed Solution

The expansion of air on opening the stopcock is sudden. The process is close to adiabatic.
$P_{1}^{- (γ - 1)} T_{0}^{γ} = P_{0}^{- (γ - 1)} T_{1}^{γ} \Rightarrow {(\frac{P_{1}}{P_{0}})}^{- (γ - 1)} = {(\frac{T_{1}}{T_{0}})}^{γ} . . . . . . (1)$

The slightly colder air inside the jar picks up heat from the surrounding and warms up to temperature T₀. The process is isochoric.
$\begin{array}{l} ∴ \frac{P_{2}}{T_{0}} = \frac{P_{0}}{T_{1}} \\ ∴ (\frac{T_{1}}{T_{0}}) = \frac{P_{0}}{P_{2}} . . . . . (2) \\ From (1) and (2) {(\frac{P_{1}}{P_{0}})}^{- (γ - 1)} = {(\frac{P_{0}}{P_{2}})}^{γ} \\ {(\frac{P_{0}}{P_{1}})}^{\frac{γ - 1}{γ}} = \frac{P_{0}}{P_{2}} \end{array}$
$\begin{array}{l} ∴ (1 - \frac{1}{γ}) \ln (\frac{P_{0}}{P_{1}}) = \ln (\frac{P_{0}}{P_{2}}) \\ \Rightarrow \ln (\frac{P_{0}}{P_{1}}) - \ln (\frac{P_{0}}{P_{2}}) = \frac{1}{γ} \ln \frac{P_{0}}{P_{1}} simplifying gives - \\ ∴ γ = \frac{\ln (\frac{P_{0}}{P_{1}})}{\ln (\frac{P_{2}}{P_{1}})} \end{array}$