Question

If 10 gm of a gas at atmospheric pressure is cooled from 273C to 0C keeping the volume constant,its pressure would become
(1)1/2 atm
(2)1/273 atm
(3)2 atm
(4)273 atm

Answer 1

For a fixed mass of gas at constant volume
P ∝ T

P1 / P2 = T1 / T2
P2 = P1 × T2 / T1
P2 = 1 atm × (0 + 273) / (273 + 273)
P2 = 1/2 atm

Pressure will become 1/2 atm
Option (1) is correct.

Answer 2

Given:

$T_1$ = 273°C = 546.15K

$P_1$= 1 atm

$T_2$ = 0°C = 273.15K

To find:

P_2 = ?

Formula to be used:

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

Calculation:

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

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

$P_2=\frac{1\times 273.15}{546.15}$

$P_2 = \frac{1}{273.1}$

Conclusion:

When 10 gm of a gas at atmospheric pressure is cooled from 273C to 0C keeping the volume constant,its pressure would become $\frac{1}{273.1}$. Thus option 2 is correct.

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