How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26ÁC? Show all work used to solve this problem

The answer is 7.33 g.

To calculate this, we will use the the ideal gas law:

PV = nRT

where

P - pressure of the gas,

V - volume of the gas,

n - amount of substance of gas,

R - gas constant,

T - temperature of the gas.

Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:

PV = RTm/M

It is given:

P = 0.98 atm

V = 10.2 l

T = 26°C = 299.15 K

R = 0.082 l atm/Kmol (gas constant)

M (H2O) = 2Ar (H) + Ar (O) = 2*1 + 16 = 2 + 16 = 18g

m = ?

Since PV = RTm/M, then:

m = PVM/RT

m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g