An ideal gas whose adiabatic exponent equals to γ=75 is expanded according to the law P=2V. The initial volume of the gas is equal to V0=1 unit. As a result of expansion, the volume increases 4 times. (Take R=253 units) Column-I Column-IIPWork done by the gas125 unitsQIncrement in internal energy of the gas290 unitsRHeat supplied to the gas375 unitsSMolar heat capacity of the gas in the process415 units

Molar heat capacity is C=CV+R1−(−1)=3R=25 unitsQ=nCΔT=n3RTf-Ti=3PfVf-PiVi ⇒Q=32VfVf-2ViVi=616-1=90 unitsΔU=nCVΔT=n52RTf-Ti=52PfVf-PiVi ⇒Q=522VfVf-2ViVi=516-1=75 unitsW=Q−ΔU=90-75=15 units

An ideal gas whose adiabatic exponent equals to γ=75 is expanded according to the law P=2V. The initial volume of the gas is equal to V0=1 unit. As a result of expansion, the volume increases 4 times. (Take R=253 units) Column-I Column-IIPWork done by the gas125 unitsQIncrement in internal energy of the gas290 unitsRHeat supplied to the gas375 unitsSMolar heat capacity of the gas in the process415 units

Molar heat capacity is C=CV+R1−(−1)=3R=25 unitsQ=nCΔT=n3RTf-Ti=3PfVf-PiVi ⇒Q=32VfVf-2ViVi=616-1=90 unitsΔU=nCVΔT=n52RTf-Ti=52PfVf-PiVi ⇒Q=522VfVf-2ViVi=516-1=75 unitsW=Q−ΔU=90-75=15 units

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An ideal gas whose adiabatic exponent equals to $γ = \frac{7}{5}$ is expanded according to the law $P = 2 V .$ The initial volume of the gas is equal to $V_{0} = 1$ unit. As a result of expansion, the volume increases 4 times. (Take $R = \frac{25}{3}$ units)
Column-I Column-II
P Work done by the gas 1 25 units
Q Increment in internal energy of the gas 2 90 units
R Heat supplied to the gas 3 75 units
S Molar heat capacity of the gas in the process 4 15 units

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$P - 1, Q - 2, R - 3, S - 4$

$P - 4, Q - 1, R - 3, S - 1$

$P - 4, Q - 3, R - 2, S - 1$

$P - 4, Q - 3, R - 1, S - 2$

answer is C.

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Detailed Solution

$Molar heat capacity is C = C_{V} + \frac{R}{1 - (- 1)} = 3 R = 25 units$

$Q = nCΔT = n 3 R (T_{f} - T_{i}) = 3 (P_{f} V_{f} - P_{i} V_{i}) \Rightarrow Q = 3 (2 V_{f} V_{f} - 2 V_{i} V_{i}) = 6 (16 - 1) = 90 units$

$Δ U = n C_{V} Δ T = n \frac{5}{2} R (T_{f} - T_{i}) = \frac{5}{2} (P_{f} V_{f} - P_{i} V_{i}) \Rightarrow Q = \frac{5}{2} (2 V_{f} V_{f} - 2 V_{i} V_{i}) = 5 (16 - 1) = 75 units$

$W = Q - ΔU = 90 - 75 = 15 units$

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