## Chapter 2

## Q. 2.31

A cylinder closed at its both ends is divided into two compartments by a frictionless and freely moving piston. Both piston and cylinder are insulated. One compartment contains air and the second one contains nitrogen. Initially both air and nitrogen have pressure 1 bar, volume 0.02 m³ and temperature 298 K. Heat is supplied from the side of air compartment by an electric heater till the volume of nitrogen is reduced to 0.01 m³ . Calculate final temperature of air and heat supplied to it. Assume for air, c_ν = 0.718 kJ/kg. K, R = 0.287 kJ/kg. K, and for Nitrogen, \gamma = 1.4.

## Step-by-Step

## Verified Solution

The set up is shown in Fig. 2.43.

The compression of nitrogen is adiabatic.

Process 1–2:

\begin{aligned}p_2 &=p_1\left\lgroup \frac{V_1}{V_2}\right\rgroup ^\gamma=1 \times\left(\frac{0.02}{0.01}\right)^{1.4}=2.639 \ bar \\ \\W_{1-2} &=\frac{p_1 V_1-p_2 V_2}{\gamma-1} \\&=10^2\left(\frac{0.02-2.639 \times 0.01}{1.4-1}\right)=-1.5975 \ kJ\end{aligned}

Process 1–3:

\begin{aligned}V_3 &=(0.02+0.02)-0.01=0.03 m ^3 \\ \\T_3 &=\frac{p_3 V_3 T_1}{p_1 V_1}=\frac{2.639 \times 0.03 \times 298}{0.02}=1179.63 \ K \\ \\m &=\frac{p_1 V_1}{R T_1}=\frac{1 \times 10^5 \times 0.02}{287 \times 298}=0.0234 \ kg \\ \\d U &=U_3-U_1=m c_ν\left(T_3-T_1\right) \\ &=0.0234 \times 0.718(1179.63-298)=14.812 \ kJ \\ \\Q_{1-3} &=W_{1-3}+d U=W_{1-2}+d U=1.5975+14.812=16.41 \ kJ\end{aligned}