## Chapter 2

## Q. 2.33

The heat capacity at constant pressure of a certain system is a function of temperature only and may be expressed as :

c_p=\frac{2.093+41.87}{(t+100)} \mathrm{J} /{ }^{\circ} \mathrm{C}

Where t is the temperature of the system in °C. The system is heated while it is maintained at a pressure of 1 atmosphere until its volume increases from 2000 cm³ to 2400 cm³ and its temperature increases from 0°C to 100°C. Calculate (a) the magnitude of heat interaction, and (b) how much does the internal energy of the system increase?

## Step-by-Step

## Verified Solution

The p–V diagram is shown in Fig. 2.45.

\begin{aligned}p_1 &=1.013 \times 10^5 \mathrm{~Pa} \\V_1 &=2 \times 10^{-3} \mathrm{~m}^3 \\V_2 &=2.4 \times 10^{-3} \mathrm{~m}^3 \\t_1 &=0^0 \mathrm{C}, t_2=100^{\circ} \mathrm{C}\end{aligned}

(a)

\begin{aligned}Q_{1-2} &=\int\limits_{t_1}^{t_2} c_p d t \\&=\int\limits_0^{100}\left[\frac{2.093+41.87}{(t+100)}\right] d t \\ \\&=|2.093 t+41.87 \operatorname{In}(t+100)|_0^{100} \\&=2.093 \times 100+41.87(\ln 200-\ln 100) \\&=209.3+41.87 \ln 2=238.32 \mathrm{~J} / \mathrm{kg}\end{aligned}

(b)

\begin{aligned}W_{1-2} &=p\left(V_2-V_1\right) \\&=1.013 \times 10^5(2.4-2) 10^{-3}=40.52 \mathrm{~J} \\Q_{1-2} &=U_{1-2}+W_{1-2} \\U_{1-2} &=238.32-40.52=197.8 \mathrm{~J}\end{aligned}