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Byju's Answer

Standard XII

Physics

Real Gas Equation

A mass m of a...

Question

A mass m of a perfect gas at pressure $p_{1}$ and volume $V_{1}$ undergoes an isothermal process. The final pressure is $p_{2}$ and volume is $V_{1}$ . The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is

p_{1} V_{1} I n \frac{V_{2}}{V_{1}}

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- p_{1} V_{1} I n \frac{p_{1}}{p_{2}}

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R T I n \frac{V_{2}}{V_{1}}

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m R T I n \frac{p_{2}}{p_{1}}

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Solution

The correct option is B $- p_{1} V_{1} I n \frac{p_{1}}{p_{2}}$
(b)
Since, according to the statement given to problem, work done on the system is considered as positive, so
$ρ W = - p d V$
(-ve sign is taken as dV is -ve during compression)
$W = - \int_{1}^{2} p d V = - \int_{1}^{2} m R T \frac{d V}{V}$
$= - m R T \int_{1}^{2} \frac{d V}{V} = - m R T \frac{V_{2}}{V_{1}}$
$= - m R T I n \frac{p_{1}}{p_{2}} = - p_{1} V_{1} I n \frac{p_{1}}{p_{2}}$

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A mass m of a perfect gas at pressure p1 and volume V1 undergoes an isothermal process. The final pressure is p2 and volume is V1. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is

The correct option is B −p1V1Inp1p2(b) Since, according to the statement given to problem, work done on the system is considered as positive, so ρW=−pdV (-ve sign is taken as dV is -ve during compression) W=−∫21pdV=−∫21mRTdVV =−mRT∫21dVV=−mRTV2V1 =−mRTInp1p2=−p1V1Inp1p2

A mass m of a perfect gas at pressure $p_{1}$ and volume $V_{1}$ undergoes an isothermal process. The final pressure is $p_{2}$ and volume is $V_{1}$ . The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is