Answer
See the detailed answer below.
Work Step by Step
a) We know that the work done on the gas is given by
$$W=-\int_i^fPdV$$
where $P$ is given by $PV=nRT$; Hence, $P=nRT/V$
$$W=-\int_{V_i}^{V_f}\dfrac{nRT}{V}dV$$
we know that the final volume is one-third the initial volume, we also know that the temperature during this process is constant.
$$W=-nRT\int_{V_i}^{\frac{1}{3}V_i}\dfrac{dV}{V}=-nRT\ln( V)\bigg|\int_{V_i}^{{\frac{1}{3}V_i}}$$
$$W=-nRT\ln\left(\dfrac{\frac{1}{3} \color{red}{\bf\not} V_i}{ \color{red}{\bf\not} V_i}\right)$$
Plugging the known;
$$W=-(2)(8.31)(30+273)\ln\left(\dfrac{1}{3}\right)$$
$$W=\color{red}{\bf 5.53\times 10^3}\;\rm J$$
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b) when the pressure is constant,
$$W=-\int_i^fPdV=-P\int_i^f dV$$
$$W=-P(\frac{1}{3}V_i-V_i)$$
$$W=\frac{2}{3}PV_i$$
where $V_i$ is given by $PV=nRT$; Hence, $V_i=nRT/P$
$$W=\frac{2}{3} \color{red}{\bf\not} P \dfrac{nRT_i}{ \color{red}{\bf\not} P}$$
$$W=\frac{2}{3}nRT_i$$
Plugging the known;
$$W=\frac{2}{3}(2)(8.31) (30+273) $$
$$W=\color{red}{\bf 3.36\times 10^3}\;\rm J$$
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c) See the graph below.
We have first an isothermal process then an isobaric process.
We need to find the final pressure of the isothermal process,
$$P_f=\dfrac{P_iV_i}{V_f}=\dfrac{3P_iV_i}{V_i}=3P_i=\bf 4.5\;\rm atm$$
