Work done by a system under isothermal change from a volume V 1 to V 2 for a gas which obeys Vander Waal-s equation V - n P - an 2- V - nRTA- n R T log 10V2-V1- n2V1-V2-V1 V2- n R T log eV2-n -V1-n - n2V1-V2-V1 V2C- n R T log eV1-n -V2-n - n2V1 V2-V1-V2n R T log eD- n R T log eV2-n -V1-n - n2V1-V2-V1 V2

The correct option is B According to given Vander Waal's equation P=nRt/V nβ α n^2/V^2 Work done, W=∫^V_2_V_1Pdv nRT ∫^V_2_V_1dV/V nβ α n^2∫^V_2_V_1dV/V^2 ...

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Byju's Answer

Standard XII

Physics

Work

Work done by ...

Question

Work done by a system under isothermal change from a volume $V_{1} t o V_{2}$ for a gas which obeys Vander Waal's equation $(V - β n) (P + \frac{a n^{2}}{V}) = n R T$

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Solution

The correct option is A
According to given Vander Waal's equation

$P = \frac{n R t}{V - n β} - \frac{α n^{2}}{V^{2}}$

Work done, $W = \int_{V_{1}}^{V_{2}} P d v - n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V - n β} - α n^{2} \int_{V_{1}}^{V_{2}} \frac{d V}{V^{2}}$

$= n R T [l o g_{e} (V - n β)]_{V_{2}}^{V_{1}} + α n^{2} {[\frac{1}{V}]}_{V_{1}}^{V_{2}}$

$= n R T l o g_{e} = \frac{V_{2} - n β}{V_{1} - n β} + α n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$

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Work done by a system under isothermal change from a volume V1 to V2 for a gas which obeys Vander Waal's equation (V−β n)(P+an2V)=nRT

The correct option is A According to given Vander Waal's equation P=nRtV−nβ−αn2V2 Work done, W=∫V2V1Pdv−nRT ∫V2V1dVV−nβ−αn2∫V2V1dVV2 =nRT[loge(V−nβ)]V1V2+αn2[1V]V2V1 =nRT loge=V2−nβV1−nβ+αn2(V1−V2V1V2)

Work done by a system under isothermal change from a volume $V_{1} t o V_{2}$ for a gas which obeys Vander Waal's equation $(V - β n) (P + \frac{a n^{2}}{V}) = n R T$