A sample of gas collected over water at 27^\circ\text{C} occupies a volume of 250\text{ cm… — JAMB UTME

A sample of gas collected over water at $27^\circ\text{C}$ occupies a volume of $250\text{ cm}^3$ at a total pressure of $750\text{ mmHg}$ . If the saturated vapor pressure of water at $27^\circ\text{C}$ is $30\text{ mmHg}$ , what is the volume of the dry gas at standard temperature and pressure (STP)?

$215.5\text{ cm}^3$ Answer
B
$224.5\text{ cm}^3$
C
$236.8\text{ cm}^3$
D
$233.5\text{ cm}^3$

Answer

The volume of the dry gas at STP is

215.5\text{ cm}^3

Subtracting the aqueous tension (

30\text{ mmHg}

) from the total pressure (

750\text{ mmHg}

) yields the true dry gas pressure (

720\text{ mmHg}

). Substituting this along with initial temperature (

300\text{ K}

) and STP conditions (

760\text{ mmHg}

and

273\text{ K}

) into the combined gas equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

gives

V_2 = 215.5\text{ cm}^3

Step-by-Step Solution

Calculate the partial pressure of the dry gas

P_{\text{dry}} = P_{\text{total}} - P_{\text{water}} = 750\text{ mmHg} - 30\text{ mmHg} = 720\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water includes the pressure of the dry gas and the saturated vapor pressure of water.

Convert temperatures to Kelvin

T_1 = 27 + 273 = 300\text{ K}

T_2 = 273\text{ K}

(STP)

Gas law equations require absolute temperature units in Kelvin.

Apply the general gas law equation to solve for the volume at STP (

V_2

)

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{720 \times 250 \times 273}{760 \times 300} = \frac{49,140,000}{228,000} = 215.5\text{ cm}^3

Standard conditions (STP) are

P_2 = 760\text{ mmHg}

and

T_2 = 273\text{ K}

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water

Estimated Time:1m 30s

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