One litre of a gas weighs 2 g at 300 K and | vedclass

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(A) Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass). Since $V$,$R$,and $M$ are constant,we have $\frac{P_1

Vedclass · Accepted Answer

$450$

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(A) Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass). Since $V$,$R$,and $M$ are constant,we have $\frac{P_1 V}{m_1 T_1} = \frac{P_2 V}{m_2 T_2}$. Rearranging for $T_2$: $T_2 = T_1 	imes \frac{P_2}{P_1} 	imes \frac{m_1}{m_2}$. Given: $P_1 = 1 \ atm$,$T_1 = 300 \ K$,$m_1 = 2 \ g$,$P_2 = 0.75 \ atm$,$m_2 = 1 \ g$. $T_2 = 300 	imes \frac{0.75}{1} 	imes \frac{2}{1} = 300 	imes 1.5 = 450 \ K$.

One litre of a gas weighs $2 \ g$ at $300 \ K$ and $1 \ atm$ pressure. If the pressure is made $0.75 \ atm$,at which of the following temperatures will one litre of the same gas weigh $1 \ g$?

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