A gas at 27 ^oC has a volume V and pressure P | vedclass

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(C) The ideal gas equation is given by $PV = nRT$, which implies $\frac{PV}{T} = 	ext{constant}$.Let the initial state be $(P_1, V_1, T_1)$ and the f

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$1527$

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(C) The ideal gas equation is given by $PV = nRT$, which implies $\frac{PV}{T} = 	ext{constant}$.Let the initial state be $(P_1, V_1, T_1)$ and the final state be $(P_2, V_2, T_2)$.Given: $T_1 = 27 ^oC = 27 + 273 = 300 K$, $P_2 = 2P_1$, and $V_2 = 3V_1$.Using the relation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$, we get:$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}ight) \left(\frac{V_2}{V_1}ight) = \left(\frac{2P_1}{P_1}ight) \left(\frac{3V_1}{V_1}ight) = 2 	imes 3 = 6$.Therefore, $T_2 = 6 	imes T_1 = 6 	imes 300 K = 1800 K$.Converting back to Celsius: $T_2 = 1800 - 273 = 1527 ^oC$.

$A$ gas at $27 ^oC$ has a volume $V$ and pressure $P$. On heating, its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be ....... $^oC$.

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