Question

A cylinder of compressed gas rolls off a boat and falls to the bottom of a lake. Eventually it rusts and the gas bubbles to the surface. A chemist collects a sample of the gas with the idea of trying to identify the gas. The wet gas collected occupies a volume of 287 mL at a pressure of 695 torr and temperature of 28.0oC. The vapor pressure of water at 28.0oC is 0.0372 atm. 1. Calculate the volume (L) that the gas occupies after it is dried (the water vapor removed) and stored at STP.

Answer 1

Answer:

0.228 L is the volume that the gas occupies after it is driedand stored at STP.

Explanation:

Pressure of the wet gas = $P=695 Torr=\frac{695}{760}atm=0.914 atm$

1 atm = 760 Torr

Pressure of water vapor = p = 0.0372 atm

Pressure of gas = $P_1=P-p=0.914 atm-0.0372 atm = 0.8768 atm$

Volume of wet gas = $V_1=287 mL=0.287 L$

1 mL = 0.001 L

Temperature of the wet gas = $T_1=28.0^oC=28.0+273 K=301 K$

Pressure of dry gas at STP = $P_2=1 atm$

Temperature of the dry gas at STP ,$T_2= 273 K$

Volume of dry gas at STP = $V_2$

Using combined gas law:

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

$\frac{0.8768 atm\times 0.287 L}{301 K}=\frac{1 atm\times V_2}{273 K}$

$V_2=0.228 L =228 mL$

0.228 L is the volume that the gas occupies after it is driedand stored at STP.