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Answer :
Final answer:
The gas collected over water at 43ºC, 596mL, and 99.1kPa will occupy a volume of approximately 533mL at a temperature of 63ºC and standard atmospheric pressure (101.3kPa).
Explanation:
A gas is collected over water and occupies a volume of 596mL at 43ºC. The atmospheric pressure is 99.1kPa. What volume will the dry gas occupy at a new temperature of 63ºC and standard atmospheric pressure (101.3kPa)? The vapor pressure of water at 43ºC is 8.6kPa.To answer this question, we can use the combined gas law which states that the ratio of the product of pressure and volume to the temperature of one state is equal to the product of pressure, volume to the temperature of another state. First, we adjust the total pressure by subtracting the water vapor pressure to obtain the dry pressure of the gas: Pd = Pt - Pv = 99.1kPa - 8.6kPa = 90.5kPa. In this state, using the combined gas law,
P1V1/T1 = P2V2/T2,
where T is in Kelvin. We plug in the initial conditions (P1 = 90.5 kPa, V1 = 596 mL, T1 = 43ºC = 316K) and the final conditions (P2 = 101.3 kPa, T2 = 63°C = 336K) and solve for V2, yielding a final volume of approximately 533 mL. Thus, after warming the gas collected over water to 63ºC and at standard atmospheric pressure, the dry volume of the gas will be around 533mL.
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