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Chemistry Question on Thermodynamics terms

The work done in adiabatic process is given by Here, symbols have their usual meaning

A

nR(T1−T2)γ\frac{nR\left(T_{1}-T_{2}\right)}{\gamma}

B

nR(T1−T2)γ−1\frac{nR\left(T_{1}-T_{2}\right)}{\gamma-1}

C

nγ(T1−T2)Rn \gamma \left(T_{1}-T_{2}\right)R

D

γ(T1−T2)Rn\frac{\gamma\left(T_{1}-T_{2}\right)R}{n}

Answer

nR(T1−T2)γ−1\frac{nR\left(T_{1}-T_{2}\right)}{\gamma-1}

Explanation

Solution

In the adiabatic process PVγ=PV^{\gamma} = constant (K)(K) If an ideal gas is changed from state (P1V1T1)(P_1V_1T_1) to state (P2V2T2)(P_2V_2T_2) adiabatically, then work done W=∫V1V2PdV=K∫V1V2dVVγW = \int\limits_{V_1}^{V_{2}} P dV = K \int_{V_1}^{V_{2}} \frac{dV}{V^{\gamma}} =K[V−γ+1−γ+1]V1V2=K1−γ[1V2γ−1−1V1γ−1]= K \left[\frac{V^{-\gamma+1}}{-\gamma +1}\right]_{V_1}^{V_{2}} = \frac{K}{1-\gamma} \left[\frac{1}{V_{2}^{\gamma-1}} -\frac{1}{V_{1}^{\gamma-1}}\right] W=11−γ[P2V2γV2γ+1−P1V1γV1γ−1]W=\frac{1}{1-\gamma} \left[\frac{P_{2}V_{2}^{\gamma}}{V_{2}^{\gamma+1}} -\frac{P_{1}V_{1}^{\gamma}}{V_{1}^{\gamma-1}}\right] =11−γ[P2V2−P1V1]= \frac{1}{1-\gamma}\left[P_{2}V_{2} -P_{1}V_{1}\right] =nR1−γ(T2−T1)=nR(T1−T2)(γ−1)= \frac{nR}{1-\gamma}\left(T_{2}-T_{1}\right) = \frac{nR\left(T_{1}-T_{2}\right)}{\left(\gamma-1\right)} [∵P1V1=nRT1andP2V2=nRT2]\quad \left[\because P_{1}V_{1} = nRT_{1} {\text{and}} P_{2}V_{2} = nRT_{2}\right]