Question #207446

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


1
Expert's answer
2021-06-16T05:32:23-0400

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 ×1molC2H428gC2H4\frac{1mol C_2H_4}{28g C_2H_4}


=0.179molC2H4=0.179mol C_2H_4

Recall the ideal gas equation

PV=nRTPV=nRT

Calculate temperature by substituting values in ideal gas equation 

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


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


T=92.9KT=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


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