Question

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

• A. $1500 \,m^3$
• B. $1700 \,m^3$
• C. $1900 \,m^3$
• D. $2700 \,m^3$

Answer 1

Answer: (D)

$V_{1} = 1500\, m^{3}, T_{1} = 27 \,{}^{\circ}C = 300\, K$
$P_{1} = 4 \,atm, T_{2} = -3 \,{}^{\circ}C = 270 \,K, \,P_{2} = 2 \,atm$
According to ideal gas equation, $\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}$
$\therefore \quad V_{2} = \frac{P_{1}V_{1}}{T_{1}}\times\frac{T_{2}}{P_{2}} = \frac{4\times1500\times270}{300\times2} = 2700\,m^{3}$