Work done per mole in an isothermal change is

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(C) For an isothermal process,the temperature $T$ remains constant.From the ideal gas equation for $1$ mole of gas,$PV = RT$,which implies $P = \frac{

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$RT \log_{e} \frac{V_2}{V_1}$

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(C) For an isothermal process,the temperature $T$ remains constant.From the ideal gas equation for $1$ mole of gas,$PV = RT$,which implies $P = \frac{RT}{V}$.The work done $W$ during an isothermal expansion from volume $V_1$ to $V_2$ is given by the integral:$W = \int_{V_1}^{V_2} P \, dV$Substituting the value of $P$:$W = \int_{V_1}^{V_2} \frac{RT}{V} \, dV$Since $R$ and $T$ are constants,we can take them out of the integral:$W = RT \int_{V_1}^{V_2} \frac{1}{V} \, dV$$W = RT [\ln V]_{V_1}^{V_2}$$W = RT (\ln V_2 - \ln V_1)$$W = RT \log_{e} \left( \frac{V_2}{V_1} \right)$.

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The volume of an ideal gas is $1 \, L$ and its pressure is equal to $72 \, cm$ of mercury column. The volume of the gas is made $900 \, cm^3$ by compressing it isothermally. The increase in the pressure of the gas will be ...... $cm$ of mercury.

During an isothermal expansion,a confined ideal gas does $-150 \, J$ of work against its surroundings. This implies that

$Assertion :$ The isothermal curves intersect each other at a certain point.
$Reason :$ The isothermal change takes place slowly,so,the isothermal curves have very little slope.

Write the first law of thermodynamics for an isothermal process in an ideal gas.

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