Answer to Question #253431 in General Chemistry for Jejemon

Question

A balloon has a volume of 4.0L when at sea level(1.0atm)at a room temperature of 28°C.What will be its volume when inflated with the same amount of a gas at an elevation where the atmospheric pressure is 700 mmhg at 28°C

Answer 1

$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}=\text{constant}$$

The balloon has a volume of 4.0 L when at sea level (1.0 atm) at a room temperature of 28 °C.

$V_1=4.0\ L=4\times 10^{-3}\ m^3,$

$P_1=1.0\ atm=101325\ Pa,$

$T_1=28\degree C=(28+273)\ K=301\ K$

1mm Hg=133.322 Pa

$P_2=700mm\ Hg=(700\cdot 133.322)\ Pa=93325.4\ Pa$

 $T_2=28\degree C=(28+273)\ K=301\ K=T_1$          

$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

Solve for $V_2$

$$V_2=\dfrac{P_1T_1}{P_2T_2}V_1$$

    

$$V_2=\dfrac{101325\ Pa(301\ K)}{93325.4\ Pa(301\ K)}\cdot4\times 10^{-3}\ m^3$$ $$\approx 4.343\times 10^{-3}\ m^3\approx4.343\ L$$

The volume of gas will be $V_2=4.343\ L.$