Question

A container is filled with gas to a pressure of 2 atm at $0^o$ C. At what temperature will the pressure inside the container be 2.50 atm?

• A. ${75}^{o}C$
• B. ${12.5}^{o}C$
• C. ${18}^{o}C$
• D. ${68.25}^{o}C$

Answer 1

The correct option is D ${68.25}^{o}C$

Answer : D. 68.25${}^{o}C$

We have ideal gas equation, $PV=nRT$

Here, only P and T changes. n, R and V (volume of the container) remain same.

From the question,

$P_1$ = 2 atm

$T_1$ = $0^o\text{C}$ = $273K$

$P_2$ = 2.50 atm

$T_2$ = ?

For $P_1$ and $T_1$ ideal gas equation becomes, $P_{1}V=nR{T}_1$ ---------> ( 1 )

For $P_2$ and $T_2$ ideal gas equation becomes, $P_{2}V=nR{T}_2$ ---------> ( 2 )

Divide equation ( 1 ) by ( 2 ) we get,

$\frac{P_{1}V}{P_{2}V}=\frac{nR{T}_1}{nR{T}_2}$

$=>\frac{P_1}{P_2}=\frac{T_1}{T_2}$ $=>\frac{2}{2.50}=\frac{273}{T_2}$

$=>T_2=273\times \frac{2.50}{2}=341.25K$

$=>T_2=341.25-{273}^{o}C={68.25}^{o}C$

option D. ${68.25}^{o}C$ is correct