Question

A 500 mL gas sample is collected over water at a pressure of 740mmHg and 25°C. What is the volume of the dry gas at STP? (STP = 1 atm and 0°C) Vapor pressure at 25° of H2O equals 24mmHg.

Answer 1

1) List the known and unknown quantities.

Sample: gas.

Volume: 500 mL.

Pressure: 740 mmHg

Temperature: 25 ºC.

Vapor pressure at 25 ºC: 24 mmHg.

2) Pressure of the gas.

$P_{gas}=P_{atm}-P_{water\text{ }vapor}$$P_{gas}=740\text{ }mmHg-24\text{ }mmHg$$P_{gas}=716\text{ }mmHg$

The pressure of the gas is 716 mmHg

3) Moles of gas

3.1- List the known quantities.

Volume: 500 mL.

Temperature: 25 ºC.

Pressure: 716 mmHg.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Set the equation.

$PV=nRT$

3.3- Convert the units of the volume, the temperature, and the pressure.

Volume.

1 L = 1000 mL

$L=500\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.500\text{ }L$

Temperature.

$K=25\text{ }ºC+273.15\text{ }K$$K=298.15\text{ }K$

Pressure

1 atm = 760 mmHg

$atm=716\text{ }mmHg*\frac{1\text{ }atm}{760\text{ }mmHg}=0.942\text{ }atm$

3.4- Plug in the know quantities in the ideal gas equation.

$(0.942\text{ }atm)(0.500\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(298.15\text{ }K)$

3.5- Solve for n (moles).

Divide both sides by (0.082057 L * atm * K^(-1) * mol^(-1)) * (298.15 K)

$\frac{(0.942atm)(0.500L)}{(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}=\frac{n(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}{(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}$$n=\frac{(0.942atm)(0.500L)}{(0.082057L*atm*K^{-1}*mol^{-1})(298.15K)}=$$n=0.0193\text{ }mol$

4) Dry volume at STP

STP conditions are

Temperature: 273 K

Pressure: 1 atm.

At STP conditions 1 mol of a gas occuppies 22.4 L. We can use this as a conversion factor.

1 mol gas = 22.4 L

$V=0.0193\text{ }mol\text{ }gas*\frac{22.4\text{ }L}{1\text{ }mol\text{ }gas}=0.432\text{ }L$

The volume of the dry gas at STP is 0.432 L.

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