Answer :
A gas, collected over water at 25°C, occupies 10.5 L. The total pressure of the collected gas is 98.5 kPa. If the gas is dried and placed in a 12.0 L flask at 30°C, the pressure of the gas will be 12.0kPa
We must take into account the vapour pressure of water at the specified temperature in order to calculate the pressure of the gas when it is dried and put in a 12.0 L flask at 30°C.
Using the vapour pressure of water at 25°C, let's determine the partial pressure of water vapour in the flask at 30°C. We can suppose that the behaviour of the water vapour is optimal.
Initial pressure of the gas (with water vapor) = 98.5 kPa
Vapor pressure of water at 25°C = 3.17 kPa
To find the partial pressure of water vapor at 30°C, we need to account for the change in temperature. For this, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔH_vap/R)((1/T1) - (1/T2))
where:
P1 = Initial pressure of water vapor
P2 = Final pressure of water vapor
ΔH_vap = Heat of vaporization of water
R = Ideal gas constant (8.314 J/(mol·K))
T1 = Initial temperature (25°C + 273.15 K)
T2 = Final temperature (30°C + 273.15 K)
The heat of vaporization of water is approximately 40.7 kJ/mol, we can substitute the values into the equation:
ln(P2/3.17 kPa) = (40.7 kJ/mol / (8.314 J/(mol·K))) * ((1 / (25°C + 273.15 K)) - (1 / (30°C + 273.15 K)))
Simplifying and solving for P2:
ln(P2/3.17 kPa) = 3.175
P2/3.17 kPa = e^3.175
P2 = 3.17 kPa * e^3.175
Pressure of the gas (dried) = Initial pressure of the gas - Partial pressure of water vapor
Pressure of the gas (dried) = 98.5 kPa - P2
Substituting the calculated value of P2 into the equation:
Pressure of the gas (dried) = 98.5 kPa - (3.17 kPa * e^3.175)=12.0kPa
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