A gas is filled in a cylinder. Its temperatur | vedclass

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(D) Using the ideal gas equation: $PV = nRT$ $(i)$Given that pressure $P$ is constant,the new volume $V_1 = V - 0.10V = 0.9V$ and the new temperature

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$25$

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(D) Using the ideal gas equation: $PV = nRT$ $(i)$Given that pressure $P$ is constant,the new volume $V_1 = V - 0.10V = 0.9V$ and the new temperature $T_1 = T + 0.20T = 1.2T$.Substituting these into the ideal gas equation for the new state: $P(0.9V) = n_1 R(1.2T)$ $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{P(0.9V)}{PV} = \frac{n_1 R(1.2T)}{nRT}$$0.9 = \frac{n_1}{n} 	imes 1.2$$\frac{n_1}{n} = \frac{0.9}{1.2} = \frac{3}{4} = 0.75$This means $n_1 = 0.75n$,which implies $75\%$ of the original gas remains.The percentage of gas that leaked out is: $\frac{n - n_1}{n} 	imes 100 = (1 - 0.75) 	imes 100 = 25\%$.

$A$ gas is filled in a cylinder. Its temperature is increased by $20\%$ on the Kelvin scale and its volume is reduced by $10\%$. How much percentage of the gas will leak out at constant pressure?

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