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\begin{document}
\title{Assignment --- First Law of Thermodynamics}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{}
\maketitle
\section*{Question 1}
First law of thermodynamics suggests that $\sum Q = \sum W$.
\begin{align*}
	Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
	75 - 40 + Q_3 &= -15 + 44 \\
	Q_3 &= -6 kJ
\end{align*}
i.e.\ from the system to the surroundings.
\section*{Question 2}
We know that,
\begin{align*}
	Q &= \Delta E + W \\
	Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
	Q &= -8600 kJ \\
	Q &= -8.6 MJ
\end{align*}
\section*{Question 3}
There is no heat transfer, therefore
\begin{align*}
	Q &=  \Delta E + W \\
	W &= - \Delta E - \Delta V \\
	W &= \int_{1}^{0} C_v \cdot dT \\
	W &= -0.718 (T_2 - T_1) \\
	W &= -50.26 \frac{kJ}{kg} \\
	\text{Total Work} &= 2 \times (-50.26) = -100 kJ
\end{align*}
\section*{Question 4}
\begin{align*}
	1000 &= a + b \times 0.2 \\
	200 &= a + b \times 1.2 \\
	b &= -800 \\
	a &= 1000 + 2 \times 800 = 1160 \\
	\therefore P &= 1160 - 800 V \\
	W &= \int_{V_1}^{V_2} P \cdot dV \\
	  &= \int_{0.2}^{1.2} (1160 - 800V)dV \\
	  &= 6000 kJ \\
	V_1 &= 1.5 \times  1000 \times \frac{0.2}{1.5} - 85 \\
		&= 215 \frac{kJ}{kg} \\
	V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
		&= 155 \frac{kJ}{kg} \\
	\Delta V &= V_2 - V_1 \\
			 &= 40 \frac{kJ}{kg} \\
	\Delta U &= m \Delta V  = 60 kJ \\
	Q &= \Delta U + W \\
	  &= 60 + 600 \\
	  &= 660 kJ \\
	U &= 1.5 PV - 85 \frac{kJ}{kg} \\
	U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
	  &= 800 V^2 - 85 \frac{kJ}{kg} \\
	\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
	\text{For maximum V,} \\
	V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
	V &= 0.725 \\
	u_{\max} = 335.5 \frac{kJ}{kg} \\
\end{align*}
\begin{align*}
	U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
\end{align*}
\section*{Question 5}
\subsection*{Part a}
\begin{align*}
	Q &= \int_{273}^{373} C_p \cdot dT \\
	t &= T - 273 K \\
	\therefore t + 100 &= T - 173 \\
	Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
	Q &= 238.32 J
\end{align*}
\subsection*{Part b}
\begin{align*}
	Q &= \Delta E + \int p\cdot dV \\
	\Delta E &= Q - p(V_2 - V_1) \\
	\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
	\Delta E &= 197.79J
\end{align*}
\end{document}