## Chapter 2

## Q. 2.35

The specific internal energy of a gas is given by the relation:

u = 3.65 pν + 100 kJ/kg

Where p is in kPa and ν in m³/kg.

3 kg and 0.30 m³ of this gas expands from an initial pressure of 600 kPa to 100 kPa according to the law given by : p V^{1 .25} = C. If the expansion is quasi – static, calculate (a) heat transferred, (b) change in internal energy and (c) work transferred.

## Step-by-Step

## Verified Solution

V_2=V_1\left\lgroup \frac{p_1}{p_2}\right\rgroup^{\frac{1}{n}}=0.3\left(\frac{600}{100}\right)^{\frac{1}{1.25}}=1.2579 \mathrm{~m}^3

(b)

\begin{aligned}d U &=U_2-U_1=3.65\left(p_2 V_2-p_1 V_1\right) \\&=3.65(100 \times 1.2579-600 \times 0.3)=-197.87 \mathrm{~kJ}\end{aligned}

(c)

\begin{aligned}W &=\int p \cdot d V=\frac{p_1 V_1-p_2 V_2}{n-1} \\&=\frac{600 \times 0.3-100 \times 1.2579}{1.25-1}=216.84 \mathrm{~kJ}\end{aligned}

(d) Q = dU + W = – 197.87 + 216.84 = 18.97 kJ