Question

The volume of a gas mixed with water vapor at 32.0C at 742torr is 1350mL. What would be the volume of the gas at 0.0C and 760torr if all the water vapor were removed? The pressure of water vapor at 32C is 36torr.

Please show me how to do this?

Answer 1

1122.5 mL

Step-by-step explanation:

In the first scenario, by Dalton's law, the total pressure is the sum of the partial pressures of the components. So, the partial pressure of the gas is:

P1 = Ptotal - Pwater = 742 - 36 = 706 torr

By the ideal gas law, the change in a state of a gas can be calculated by:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is the temperature in K, 1 is the state 1, and 2 the state 2.

P1 = 706 torr, V1 = 1350 mL, T1 = 32ºC + 273 = 305K

P2 = 760 torr, T2 = 0ºC + 273 = 273 K

706*1350/305 = 760*V2/273

760V2/273 = 3124.92

760V2 = 853102.62

V2 = 1122.5 mL