Answer to Question #207446 in Physical Chemistry for Avi

Question

5.0 g of hydrocarbon gas formula C2H4 has a volume of 1.3 L at 1.05 atm. The temperature of the gas is

Answer 1

Given mass of hydrocarbon C₂H₄ is 5.0 g Given volume of hydrocarbon is 1.3 L and pressure is 1.05 atm.

Molar mass of C₂H₄ is 28 g/mol

We have to calculate the temperature of the hydrocarbon.

 Convert 5.0g C₂H₄ to corresponding mole:-

5.0g C₂H₄ =5.0g C₂H₄ ×$\frac{1mol C_2H_4}{28g C_2H_4}$

$=0.179mol C_2H_4$

Recall the ideal gas equation

$PV=nRT$

Calculate temperature by substituting values in ideal gas equation 

$T=\frac{PV}{nR}$

$T=\frac{(1.05atm)×(1.3L)}{(0.179mol)×(0.08206L/atm/mol/K)}$

$T=92.9K$

Therefore , temperature of gas is 92.9K

Therefore the temperature in celsius scale is = (92.9-273) °C

  = -180.1 °C 

Hence temperature of the gas is -180.1°C