Question

By what fractions the original volume should be multiplied if conditions were changed to $0^{\circ }C$ and 0.92 atmosphere pressure of the original gas at ${20}^{\circ }C$ and 1 atmosphere pressure?

• A. $\frac{293}{273}\times \frac{1}{0.92}$
• B. $\frac{273}{293}\times \frac{0.92}{1}$
• C. $\frac{273}{293}\times \frac{1}{0.92}$
• D. $\frac{293}{273}\times \frac{0.92}{1}$

Answer 1

Answer: (C)

$\frac{273}{293}\times \frac{1}{0.92}$

By the fraction $\frac{273}{293}\times \frac{1}{0.92}$, the original volume should be multiplied if conditions were changed to 0 ${}^{\circ }C$ and 0.92 atmosphere pressure of the original gas at 20 ${}^{\circ }C$ and 1 atmosphere pressure.
Initial temperature $T=20$${}^{\circ }C$ = $293K$
Final temperature $T^{\prime }=0$${}^{\circ }C$ = $273K$
Initial pressure $P=1atm$
Final pressure $P^{\prime }=0.92atm$.
Initial volume $=V$
Final volume $=V^{\prime }$
According to ideal gas equation, for a fixed mass of a gas,
$\frac{V^{\prime }}{V}=\frac{T^{\prime }}{T}\times \frac{P}{P^{\prime }}\frac{V^{\prime }}{V}=\frac{273\times }{293}\times \frac{1}{0.92}{V}^{\prime }=V\times \frac{273}{293}\times \frac{1}{0.92}$