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(A) Using the ideal gas equation: $PV = nRT$,where $n = \frac{w}{M}$.So,$PV = \frac{w}{M}RT$,which implies $\frac{PV}{wT} = \frac{R}{M} = 	ext{consta

Vedclass · Accepted Answer

$450$

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(A) Using the ideal gas equation: $PV = nRT$,where $n = \frac{w}{M}$.So,$PV = \frac{w}{M}RT$,which implies $\frac{PV}{wT} = \frac{R}{M} = 	ext{constant}$.Therefore,$\frac{P_1 V_1}{w_1 T_1} = \frac{P_2 V_2}{w_2 T_2}$.Given: $P_1 = 1 \ atm, V_1 = 1 \ L, w_1 = 2 \ g, T_1 = 300 \ K$.Given: $P_2 = 0.75 \ atm, V_2 = 1 \ L, w_2 = 1 \ g, T_2 = ?$.Substituting the values: $\frac{1 	imes 1}{2 	imes 300} = \frac{0.75 	imes 1}{1 	imes T_2}$.$T_2 = 0.75 	imes 2 	imes 300 = 450 \ K$.

At $300 \ K$ temperature and $1 \ atm$ pressure,the weight of $1 \ L$ of a gas is $2 \ g$. If the pressure is changed to $0.75 \ atm$,at what temperature in $K$ will the weight of $1 \ L$ of the same gas be $1 \ g$?

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