Question

The temperature of an ideal gas at atmospheric pressure is $300K$ and volume $1m^3$.If temperature and volume become double, then pressure will be:

• A. $10^{5}N/m^2$
• B. $2\times 10^{5}N/m^2$
• C. $0.5\times 10^{5}N/m^2$
• D. $4\times 10^{5}N/m^2$

Answer 1

Answer: (A)

From ideal gas equation
$PV=nRT$
$\Rightarrow PV\propto T$
$\therefore \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\Rightarrow P_2=\frac{P_1{V}_1{T}_2}{V_2{T}_1}$
Here $:P_1=1\times 10^{5}N/m^2,$
$V_1=1m^3$
$T_1=300K,$
$V_2=2$
$V_1=2m^3$
$T_2=2$
$T_2=600K$
Putting given values in eq. (1)
$P_2=\frac{1\times 10^5\times 1\times 600}{2\times 300}$
$=10^{5}N/m^2$