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Standard XII

Chemistry

PV Work

The work done...

Question

The work done y a gas is an isothermal change where 1 refers to initial state and 2 refers to final state is

μ R (T_{2} - T_{1}) ln (\frac{V_{2}}{V_{1}})

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μ R T_{1} ln (\frac{V_{2}}{V_{1}})

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μ R T_{2} ln (\frac{V_{1}}{V_{2}})

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μ R (\frac{T_{1} + T_{2}}{2}) [ln V_{2} - ln V_{1}]

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Solution

The correct option is B $μ R T_{1} ln (\frac{V_{2}}{V_{1}})$
$w = - \int P d V P V = n R T$
$T \to c o n s t$
$= - \int_{v_{1}}^{v_{2}} \frac{n R T}{V} d V$
$= μ R T_{1} l n (\frac{V_{2}}{V_{1}})$

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The work done y a gas is an isothermal change where 1 refers to initial state and 2 refers to final state is

The correct option is B μRT1ln(V2V1)w=−∫PdVPV=nRTT→const=−∫v2v1nRTVdV=μRT1ln(V2V1)

The correct option is B $μ R T_{1} ln (\frac{V_{2}}{V_{1}})$
$w = - \int P d V P V = n R T$
$T \to c o n s t$
$= - \int_{v_{1}}^{v_{2}} \frac{n R T}{V} d V$
$= μ R T_{1} l n (\frac{V_{2}}{V_{1}})$