Question

The temperature of an ideal gas at atmospheric pressure is $300\,K$ and volume $1\,m^3$.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
$P V=n R T $
$\Rightarrow P V \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}=1 \, m ^{3} $
$T_{1}=300 \,K ,$
$ V_{2}=2 $
$V_{1}=2 \,m ^{3} $
$T_{2}=2 $
$T_{2}=600 \, K$
Putting given values in eq. (1)
$P_{2}=\frac{1 \times 10^{5} \times 1 \times 600}{2 \times 300} $
$=10^{5} \, N / m ^{2}$