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In carnot heat engine,is work done from A to B = RT1 log V2/V1 ?or nRT1 log V2/V1please answerphysics question.​

Answer»

In case of one MOLE of an IDEAL gas, the WORK DONE during isothermal process is,

W=RT_1\log\left (\dfrac {V_2}{V_1}\right)

In case of n moles of the gas, the work done is,

W=nRT_1\log\left (\dfrac {V_2}{V_1}\right)

Usually first case, i.e., one mole of gas is considered.



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