Find the mass of 22.7 liters of CH4 gas at 200.0 kPa and 50.0°C.

Mass = 27.2 g

Explanation:

Given data:

Volume of gas = 22.7 L

Pressure of gas = 200.0 KPa (200/101=1.97 atm)

Temperature = 50.0°C

Mass of gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

50+273 = 323K

1.97 atm ×22.7 L = n × 0.0821 atm.L/ mol.K  × 323K

44.72 atm.L = n × 26.52 atm.L/ mol

n =  44.72 atm.L / 26.52 atm.L/ mol

n = 1.7 mol

Mass of CH₄:

Mass =  number of moles × molar mass

Mass = 1.7 mol × 16 g/mol

Mass = 27.2 g

14 days is the half-life of phosphorus-32

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