First slide

Heat ;work and internal energy

Question

A sample of ideal gas is expanded to twice its original volume of $1 m^{3}$ in quasistatic process for which $P = α V^{2}$ with $α = 3 \times 10^{5} P a / m^{6}$ as shown in figure. Work done by the expanding gas is

Moderate

A

$7 \times 10^{5} J$
B

$3 \times 10^{5} J$
C

$6 \times 10^{5} J$
D

$8 \times 10^{5} J$

Solution

Work done, $W = \int_{V_{1}}^{V_{2}} α V^{2} d V$
$\begin{array}{l} W = {[\frac{α V^{3}}{3}]}_{1}^{2} = \frac{3 \times 10^{5}}{3} \times [2^{3} - 1^{3}] \\ W = 7 \times 10^{5} J \end{array}$

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