Answer to Question #197502 in Software Engineering for archie

Question

A balloon has a volume of 4.0 L when at sea level (1.0 atm) at room temperature 0f 28 . C What will be its volume when inflated with inflated with the same amount of gas at an elevation where the atmospheric pressure is 700 mm Hg at 28 . C

Answer 1

The ideal-gas equation

$$PV=nRT$$

The amount of the gas is the same.

Then

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

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

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

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

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

$1 mm\ Hg = 133.322\ Pa$

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

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

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

Solve for $V_2$

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

Substitute

    

$$V_2=\dfrac{101325\ Pa(301\ K)}{93325.4\ Pa(301\ K)}\cdot4\times 10^{-3}\ m^3$$

$$\approx 4.343\times 10^{-3}\ m^3\approx4.343\ L$$

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