If one mole of ideal gas|KTG & THERMODYNAMICS|JEEMAIN 2021

JEE MAIN 2021

24-02-2021 S2

Question

If one mole of an ideal gas at $\left(P_1, V_1\right)$ is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value $(B \rightarrow C)$. Then it is restored to its initial state by a reversible adiabatic compression ( C to A ). The net workdone by the gas is equal to:

Select the correct option:

$\mathrm{RT}\left(\ln 2-\frac{1}{2(\gamma-1)}\right)$

$-\frac{\mathrm{RT}}{2(\gamma-1)}$

$0$

$R T \ln 2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Ditto, word-to-word, exactly as written in the image: A – B = isothermal process W_AB = P₁V₁ ln [ 2V₁ / V₁ ] = P₁V₁ ln(2) B – C → Isochoric process W_BC = 0 C – A → Adiabatic process W_CA = (P₁V₁ − 1/4 × 2V₁) / 1 − γ = P₁V₁ [ 1 − 1/2 ] / 1 − γ = P₁V₁ / 2(1 − γ) W_net = W_AB + W_BC + W_CA = P₁V₁ ln(2) + P₁V₁ / 2(1 − γ) {P₁V₁ = RT} W_net = RT [ ln(2) − 1 / 2(γ − 1) ]

Question Tags

JEE Main

Physics

Medium

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