Question

If 10 g of a gas at atmospheric pressure is cooled from ${273}^{\circ }C$ to $0^{\circ }C$, keeping the volume constant, its pressure would become :

• A. 1/273 atm
• B. 2 atm
• C. $\frac{1}{2}atm$
• D. $5.05\times {10}^{4}N{m}^{-2}$

Answer 1

Answer: (C), (D)

$\frac{1}{2}atm$
$5.05\times {10}^{4}N{m}^{-2}$

Given,

$T_1={273}^{\circ }$ or 546.15 K
$P_1=1$ atm
$T_2=0^{\circ }C$ or 273.15 K
$P_2=?$

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\therefore P_2=\frac{P_1\times T_2}{T_1}=\frac{1\times 273.15}{546.15}$

$=\frac{1}{2}atm$ or $5.05\times {10}^{4}N{m}^{-2}$

Hence, the correct options are $\text{C}$ and $\text{D}$