Question

Chlorodifluoromethane (CHF₂Cl) was widely used in the compression/cooling circuits of refrigeration or air-conditioning systems. Since the discovery that such compounds (HCFCs and CFC's) released into the atmosphere were a major cause of depletion of stratospheric ozone, newer refrigeration systems make use of certain hydrofluorocarbons (HFCs), which are degraded in the lower atmosphere, instead. Often, mixtures of such compounds are used.

Suppose a sample of refrigerant gas consisting of a simple mixture of the gases pentafluoroethane (C₂HF₅) and difluorormethane (CH₂F₂) has a density of 1.79 g/L at 40. °C and 0.520 atm.

Calculate the volume percentage of CH₂F₂ in the sample.

Answer 1

Sol.

Volume of sample = 1 L

Density of sample = 1.79 g / L

Mass of sample = 1.79 g / L × 1 L = 1.79 g

Let the volume of CH2F2 be V1  

So , volume of C2HF5 = V2 = 1 - V1   

Pressure = P = 0.520 atm

Temperature = T = 40 °C = 40 + 273.15 K = 313.15 K

Gas constant = R = 0.0821 L atm / K mol  

Molar mass of CH2F2 = M1 = 52.024 g/mol

Molar mass of C2HF5 = M2 = 120.02 g/mol

As Mass of sample

= mass of CH2F2 + mass of C2HF5

= ( M1 × P × V1 / ( R × T ) ) + ( M2 × P × V2 / ( R × T ) )

= ( P / ( R × T ) ) × ( ( M1 × V1 ) + ( M2 × V2 ) )

So , 1.79 = ( 0.520 / ( 0.0821 × 313.15 ) ) × ( ( 52.024 × V1 ) + ( 120.02 × V2 ) )

1.79 = 0.0202258 × ( ( 52.024 × V1 ) + ( 120.02 × ( 1 - V1 ) ) )

52.024 × V1 + 120.02 - 120.02 × V1 = 88.50082  

67.996 × V1 = 31.51918  

V1 = 31.51918 / 67.996 = 0.464 L  

Therefore , volume percentage of CH2F2 in the sample

= ( volume of CH2F2 , V1 / volume of sample ) × 100

= ( 0.464 L / 1 L ) × 100

=   46.4 % ( answer )