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Number of moles of an ideal gas$=n$

Temperature$=T_1$

Volume$=V_1$

Since the process is isothermic, the final temperature will be equal to the initial one,$T_1.$

Hence, the initial temperature is equal to the final temperature.

Number of moles of an ideal gas $=n$

Temperature $=T_1$

Volume$=V_1$

(b) The work in an isothermic process is given by

$W=-nRT\ln \left(\frac{V_2}{V_1}\right)$

Since the volume was doubled, in our case we simply have

$W=-nR{T}_1\ln \left(2\right)$

Hence, the work done on the gas is$W=-nR{T}_1\ln \left(2\right).$

Number of moles of an ideal gas $=n$

Temperature $=T_1$

Volume $=V_1$

(c) Since the process is isothermal, the internal energy of the gas didn't change.

This means that, from the first law, the entire heat added to it was converted into work. Therefore, the heat added to the gas was

$Q=nR{T}_1\ln \left(2\right)$

Hence, the heat energy transferred to the gas is$Q=nR{T}_1\ln \left(2\right).$