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

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(C) Using the ideal gas law equation for a fixed amount of gas: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given initial conditions: $P_1 = P$,$V_1 =

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$1527$

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(C) Using the ideal gas law equation for a fixed amount of gas: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given initial conditions: $P_1 = P$,$V_1 = V$,$T_1 = 27^oC = 27 + 273 = 300 K$.Given final conditions: $P_2 = 2P$,$V_2 = 3V$.Substituting these values into the equation:$\frac{P 	imes V}{300} = \frac{(2P) 	imes (3V)}{T_2}$.Simplifying the equation:$\frac{1}{300} = \frac{6}{T_2}$.$T_2 = 6 	imes 300 = 1800 K$.To convert the temperature to Celsius: $T(^oC) = T(K) - 273 = 1800 - 273 = 1527^oC$.

$A$ 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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