If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

Question

Smart Achievers · Accepted Answer

(3)

$I. W_{1 \to 2 \to 3} = W_{1 \to 2} + W_{2 \to 3}$

$= n R T \ln (\frac{V_{f}}{V_{i}}) + P d V$

$= 1 R \frac{T_{0}}{3} \ln (\frac{2 V_{0}}{V_{0}}) + zero$ $[∵ n = 1, and W_{2 \to 3} isochoric no change in volume \Rightarrow Δ V = 0]$

$= \frac{R T_{0}}{3} \ln 2$ $∴$ $I \to P$

$II. Δ U_{1 \to 2 \to 3} = Δ U_{1 \to 2} + Δ U_{2 \to 3}$

$= 0 + n C_{v} Δ T = n \frac{f}{2} R Δ T$ $[Process 1 \to 2 isothermal, Δ T = 0]$

$= 1 \times \frac{3 R}{2} (T_{0} - \frac{T_{0}}{3}) = R T_{0}$ $[For monoatomic gas f = 3]$

$∴$ $II \to R$

$III. Q_{1 \to 2 \to 3} = W_{1 \to 2 \to 3} + Δ U_{1 \to 2 \to 3}$

$From first law of thermodynamics$

$= \frac{R T_{0}}{3} \ln 2 + R T_{0} = \frac{R T_{0}}{3} [\ln 2 + 3]$

$∴$ $III \to T$

$IV. Q_{1 \to 2} = W_{1 \to 2} + Δ U_{1 \to 2}$

$= \frac{R T_{0}}{3} \ln 2 + 0 = \frac{R T_{0}}{3} \ln 2$ $[∵ Δ U_{1 \to 2} = 0]$

$∴$ $IV \to P$

Solution

Q.
If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

1 I → P, II → T, III → Q, IV → T

2 I → S, II → T, III → Q, IV → U

3 I → P, II → R, III → T, IV → P

4 I → P, II → R, III → T, IV → S

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