A sample of O_2 gas has a volume of 100 ml a | vedclass

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(D) Given: Initial pressure $P_1 = 1 \ atm$,Initial volume $V_1 = 100 \ ml$,Initial temperature $T_1 = 27 + 273 = 300 \ K$.Final pressure $P_2 = 2 \ a

Vedclass · Accepted Answer

$327$

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(D) Given: Initial pressure $P_1 = 1 \ atm$,Initial volume $V_1 = 100 \ ml$,Initial temperature $T_1 = 27 + 273 = 300 \ K$.Final pressure $P_2 = 2 \ atm$,Final volume $V_2 = 100 \ ml$.Since the volume remains constant $(V_1 = V_2)$,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Substituting the values: $\frac{1}{300} = \frac{2}{T_2}$.Solving for $T_2$: $T_2 = 300 	imes 2 = 600 \ K$.Converting to Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.

$A$ sample of $O_2$ gas has a volume of $100 \ ml$ at a pressure of $1 \ atm$ and a temperature of $27^{\circ}C$. If the pressure is increased to $2 \ atm$ while the volume remains $100 \ ml$,the new temperature will be .... $^{\circ}C$.

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