A fluid undergoes a reversible adiabatic compression from 0.5 MPa, 0.2 m $^{3}$ to 0.05 m $^{3}$ according to the law PV $^{1.3}$ = constant. The change in internal energy during the process is _______ kJ .

120, kJ

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171.9, kJ

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571.7, kJ

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271.9, kJ

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Solution

The correct option is B $171.9, kJ$
Given; $P_{1} = 0.5 M P a; V_{1} = 0.2 m^{3}; V_{2} = 0.05 m^{3}$

From the relation,

$P_{1} V_{1}^{1.3} = P_{2} V_{2}^{1.3}$

$P_{2} = \frac{P_{1} V_{1}^{1.3}}{V_{2}^{1.3}} = 0.5 \times {(\frac{0.2}{0.05})}^{1.3}$

$P_{2} = 3.0314 M P a$

$N o w, T d S = d H - V d P$

For reversible adiabatic process,

[ $∵$ dS = 0]

dH = VdP

$A l s o, P_{1} V_{1}^{n} = P V^{n}$

$V = {(\frac{P_{1} V_{1}^{n}}{P})}^{\frac{1}{n}}$

$\int_{H_{1}}^{H_{2}} d H = \int_{P_{1}}^{P_{2}} V d P$

$H_{2} - H_{1} = \int_{P_{1}}^{P_{2}} {(\frac{P_{1} V_{1}^{n}}{P})}^{\frac{1}{n}} d P = \frac{n (P_{2} V_{2} - P_{1} V_{1})}{n - 1}$

$= \frac{1.3 (3031.4 \times 0.05 - 500 \times 0.2)}{1.3 - 1} = 223.47 k J$

Also,
$H_{2} - H_{1} = (U_{2} - U_{1}) + (P_{2} V_{2} - P_{1} V_{1})$

$∴$ $U_{2} - U_{1} = (H_{2} - H_{1}) + (P_{2} V_{2} - P_{1} V_{1})$

$= 223.47 - 51.57 = 171.9 k J$

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