Question

A balloon contains $1500{\text{m}}^3$ of helium gas at $27{}^{\circ }\text{C}$ and $4\text{atm}$ . The volume of helium gas at $-3{}^{\circ }\text{C}$ and $2\text{atm}$ will be

• A. $2700{\text{m}}^3$
• B. $1900{\text{m}}^3$
• C. $1700{\text{m}}^3$
• D. $1500{\text{m}}^3$

Answer 1

The correct option is A $2700{\text{m}}^3$

Given:
Initial volume, $V_1=1500{\text{m}}^3$
Initial temperature, $T_1=27{}^{\circ }\text{C}=300\text{K}$
Initial pressure, $P_1=4\text{atm}$
Final temperature, $T_2=-3{}^{\circ }\text{C}=270\text{K}$
Final pressure, $P_2=2\text{atm}$
To find:
Final volume, $V_2=?$

We know that, ideal gas equation is given by
$PV=nRT$
As number of moles will remain constant, so
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\Rightarrow V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}$
$\Rightarrow V_2=\frac{4\times 1500\times 270}{2\times 300}$
$\Rightarrow V_2=2700{\text{m}}^3$