1.00 mol of a van der Waals gas is compressed from 20.0 dm3 to 10 dm3 at 300K. I

Question

1.00 mol of a van der Waals gas is compressed from 20.0 dm3 to 10 dm3 at 300K. In the process, 20.2 kJ of work is done on the gas. Given that ?= [(2a/RT)-b]/Cp,m, with Cp,m = 38.4 J K-1 mol-1 , a = 3.60 dm6 atm mol-2 , and b=0.044 dm3 mol-1 , calculate H for the process 1.00 mol of a van der Waals gas is compressed from 20.0 dm3 to 10 dm3 at 300K. In the process, 20.2 kJ of work is done on the gas. Given that ?= [(2a/RT)-b]/Cp,m, with Cp,m = 38.4 J K-1 mol-1 , a = 3.60 dm6 atm mol-2 , and b=0.044 dm3 mol-1 , calculate H for the process 1.00 mol of a van der Waals gas is compressed from 20.0 dm3 to 10 dm3 at 300K. In the process, 20.2 kJ of work is done on the gas. Given that ?= [(2a/RT)-b]/Cp,m, with Cp,m = 38.4 J K-1 mol-1 , a = 3.60 dm6 atm mol-2 , and b=0.044 dm3 mol-1 , calculate H for the process

Explanation / Answer

µ = {(2a/RT) b}/Cp,m, with Cp,m = 38.4 J K1 mol1, a = 3.60 dm6 atm mol-2 and

b=0.044 dm3 mol-1, H for the process = ?

We can calculate µ using aboves values and also H can be calculated using dH = Cp (T2 - T1)

dH = 38.4 (300 - 273)

dH = 1036.8 J/Kg

Also µ = { ( 2 x 3.60 / 0.0821 x 300) - 0.044 } / 38.4

= 0.006466

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