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 V 2 . 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

Correct Answer is : −P1V1ln⁡P1P2

P_{1}V_{1}\ln ⁡ {V_{2}}/{V_{1}}

−P_{1}V_{1}\ln ⁡ {P_{1}}/{P_{2}}

GATE Mechanical Engineering (ME) 2017 Shift 2 Solved Paper

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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 V₂. 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

$P_{1}V_{1}\ln ⁡\;{V_{2}}/{V_{1}}$
$−P_{1}V_{1}\ln ⁡\;{P_{1}}/{P_{2}}$
$RT\ln ⁡\;{V_{2}}/{V_{1}}$
$−mRT\ln ⁡\;{P_{2}}/{P_{1}}$

Validate

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