Question

A gas occupies a volume of $0.35{dm}^3$ at $290K$ and $92.458KN{m}^{-2}$ pressure. Calculate the volume of gas at $STP$:

• A. $0.3d{m}^3$
• B. $0.15d{m}^3$
• C. $0.6d{m}^3$
• D. $0.45d{m}^3$

Answer 1

The correct option is A $0.3d{m}^3$

Given : $V_1=0.35$ $d{m}^3$ at $290K$

$P_1=92.458KN{m}^{-2}$

So, let's say that for the initial state of the gas, $P_1{V}_1=n\times R{T}_1⟶\left(1\right)$

For the second state of gas, i.e. at STP, $P_2{V}_2=n\times R{T}_2⟶\left(2\right)$

on dividing eq( 1 )and eq (2)we get,

$=\frac{P_1{V}_1}{P_2{V}_2}=\frac{nR{T}_1}{nR{T}_2}\iff \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

At STP, $P_2=100KPa{T}_2=0°C$

$⟹V_2=\frac{P_1}{P_2}\times \frac{T_2}{T_1}\times V_1$

$V_2=\frac{92.558\times {10}^{3}N{m}^{-2}}{100\times {10}^{3}N{m}^{-2}}\times \frac{(273.15+0)K}{290K}\times 0.35$ $d{m}^3=0.3051$ $d{m}^3$

Hence volume is $0.31$ $d{m}^3$.