How to calculate the pressure of an ideal gas that undergoes isothermal expansion?

Question

A 2.35 mole sample of an ideal gas, for which $C_{\mathrm{m},v}=3R/2$ initially at $\pu{27 ^\circ C}$ and $\pu{1750 kPa}$, undergoes a two stage transformation. For each of the stages described in the following list, calculate the final pressure as well as $q$, $w$, $\Delta U$ and $\Delta H$. Also calculate $q$, $w$, $\Delta U$ and $\Delta H$ for the entire process.

(a) The gas is expanded isothermally and reversibly until the volume triples.
(b) Beginning at the end of the first stage, the temperature is raised to $\pu{105 ^\circ C}$ at constant volume.

Since the process is isothermal, $\Delta H = 0$. In order to calculate the final pressure. I use the formula, $P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$, where $x = C_{\mathrm{m},p}/C_{\mathrm{m},v}$. Hence $x = 5/3$. I then substitute this into the formula to get $$P_\mathrm{i}V_\mathrm{i}^{5/3} = P_\mathrm{f}V_\mathrm{f}^{5/3}.$$ Since the volume triples, $V_\mathrm{f}/V_\mathrm{i}=3$, hence manipulating the formula I got $$P_\mathrm{i}/P_\mathrm{f}=3^{5/3}$$ and since $P_\mathrm{i}=\pu{1750 kPa}$, I substitute this into the above equation to get $P_\mathrm{f}$, which answers the first part of the question.

However I don't seem to get the right answer, hence I could not continue doing the question. Could anyone explain?

Also, am I right to say that we use $P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$ for an reversible adiabatic process only. $w = -nRT\ln\left(T_\mathrm{f}/T_\mathrm{i}\right)$ for an isothermal process and $\Delta U = nC_{\mathrm{m},v}(T_\mathrm{f} - T_\mathrm{i}) = -P_{\text{external}}(V_\mathrm{f} - V_\mathrm{i})$ for a non-reversible adiabatic process.

Im rather confused as to decide under which conditions should I use each of these three formulas.

Answer 1

I am going to use the physics conventions of work according to which, work done by the gas is positive. So if you want the answers as per chemistry conventions than just flip the sign of the work done, keeping the magnitude same.

Part (a): Reversible Isothermal Expansion

\begin{align} \Delta U_a &=nC_v\Delta T\\ \Delta U_a &=\pu{0 J}\\ \end{align}

\begin{align} \Delta H_a &=nC_p\Delta T\\ \Delta H_a &=\pu{0 J}\\ \end{align}

\begin{align} W_a&= \int_{V_i}^{V_f} P.dV\\ W_a&= \int_{V_i}^{V_f} \frac{nRT}{V}.dV\\ W_a&= nRT\ln\frac{V_f}{V_i}\\ W_a&= \pu{2.35 \times 8.314 \times 300 \times \ln3 J}\\ W_a&= \pu{6439.37 J}\\ \end{align}

\begin{align} q_a&= W_a+ \Delta U_a\\ q_a&= \pu{(6439.37 + 0)J}\\ q_a&= \pu{6439.37 J}\\ \end{align}

By Boyle's Law,

\begin{align} P_iV_i&=P_fV_f\\ P_f&=P_i\times \frac{V_i}{V_f}\\ P_f&= \pu{\frac{1750}{3} kPa}\\ P_f&= \pu{583.33 kPa}\\ \end{align}

Part (b): Isochoric Heating

\begin{align} \Delta U_b &=nC_v\Delta T\\ \Delta U_b &=\pu{2.35 \times 1.5 \times 8.314 \times 78 J}\\ \Delta U_b &=\pu{2285.93 J}\\ \end{align}

\begin{align} \Delta H_b &=nC_p\Delta T\\ \Delta H_b &=\pu{2.35 \times 2.5 \times 8.314 \times 78 J}\\ \Delta H_b &=\pu{3809.89 J}\\ \end{align}

\begin{align} W_b&= \int_{V_i}^{V_f} P.dV\\ W_b&= \pu{0 J}\\ \end{align}

\begin{align} q_b&= W_b+ \Delta U_b\\ q_b &= \pu{(0 + 2285.93)J}\\ q_b&= \pu{2285.93 J}\\ \end{align}

By Gay-Lussac's Law, \begin{align} \frac{P_i}{T_i}&=\frac{P_f}{T_f}\\ P_f&=P_i\times \frac{T_f}{T_i}\\ P_f&= \pu{\frac{1750}{3} \times \frac{378}{300} kPa}\\ P_f&= \pu{735 kPa}\\ \end{align}

Overall Process:

\begin{align} \Delta U&= \Delta U_a+ \Delta U_b\\ \Delta U &= \pu{(0 + 2285.93)J}\\ \Delta U&= \pu{2285.93 J}\\ \end{align}

\begin{align} \Delta H&= \Delta H_a+ \Delta H_b\\ \Delta H&= \pu{(0 + 3809.89)J}\\ \Delta H&= \pu{3809.89 J}\\ \end{align}

\begin{align} W&= W_a+ W_b\\ W&= \pu{(6439.37 + 0)J}\\ W&= \pu{6439.37 J}\\ \end{align}

\begin{align} q&= q_a+ q_b\\ q&= \pu{(6439.37 + 2285.93)J}\\ q&= \pu{8725.30 J}\\ \end{align}