Temperature of 2 moles of a monoatomic gas is increased from T to 2T. Match the following two columns.

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Solution

(A) Work done in isobaric process is $P Δ V = n R Δ T = 2 R Δ T$
(B) Change in Internal energy in isobaric process is $n C_{V} d t = 2 \times 3 R Δ T / 2 = 3 R Δ T$
(C) Work done in adiabatic process is $\frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = - 2 R Δ T / \frac{2}{3} = - 3 R T$
(B)Change in Internal energy in adiabatic process is $n C_{V} d t = 3 R Δ T$

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Temperature of 2 moles of a monoatomic gas is increased from T to 2T. Match the following two columns.

(A) Work done in isobaric process is PΔV=nRΔT=2RΔT(B) Change in Internal energy in isobaric process is nCVdt=2×3RΔT/2=3RΔT(C) Work done in adiabatic process is P1V1−P2V2γ−1=−2RΔT/23=−3RT(B)Change in Internal energy in adiabatic process is nCVdt=3RΔT