Question

A cylinder of ideal gas is closed by an 8 kg movable piston of area $60c{m}^2$. The atmospheric pressure is 100 kPa. When the gas is heated from ${30}^{\circ }C$ to ${100}^{\circ }C$ the piston rises 20 cm. The piston is then fastened in the place and the gas is cooled back to ${30}^{\circ }C$. If $△Q_1$ is the gas during heating and $△Q_2$is the heat lost during cooling, then the difference is $(130+x)$. Find the value of $x$.

Answer 1

Total pressure, $P={10}^5+\frac{8\times 9.8}{60\times {10}^{-4}}=1.13\times {10}^5\frac{N}{m^2}$

$\Delta V=0.2\times 60\times {10}^{-4}{m}^3$

$\Delta Q_1=\Delta U_1+\Delta W_1=\Delta U_1+p\Delta V=\Delta U_1+136J$

$\Delta U_1=nC({100}^{o}C-{30}^{o}C)$

$\Delta Q_2=\Delta W_2+\Delta U_2=0+nC({30}^{o}C-{100}^{o}C)$

$\Rightarrow \Delta Q_1-\Delta Q_2=136J$