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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}

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}

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} $

$ \text{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} $

$ \text{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} $

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}

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} $

$ \text{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} $

$ \text{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} $

Hope this helps!

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} $

$ \text{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} $

$ \text{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} $