Answer
See the detailed answer below.
Work Step by Step
a) We know that the gas undergoes an isothermal expansion process which means that
$$\boxed{T_2=T_1}$$
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b) The work done on the gas is given by
$$W=-\int_1^2 PdV$$
we know that and that $V_2=2V_1$ and that $P =nRT_1 /V $, so
$$W=-\int_{V_1}^{2V_1} \dfrac{nRT}{V}dV$$
where $nRT$ is constant
$$W=-nRT_1\int_{V_1}^{2V_1} \dfrac{dV}{V}=-nRT_1\ln\left[\dfrac{2V_1}{V_1}\right]$$
$$\boxed{W=-nRT_1\ln[2]}$$
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c) The heat energy transferred is given by
$$\Delta E_{th}=W+Q=mc\Delta T $$
and $\Delta T=0$ since it is an isothermal process, thus
$$Q=-W $$
Plugging from above;
$$\boxed{Q=nRT_1\ln[2]}$$
