Question

A quantity of nitrogen gas in a piston–cylinder assembly undergoes a process at a constant pressureof 80 bar from 220 to 300 K. Determine the work and heat transfer for the process, each in kJ per kmol of nitrogen.

Answer 1

Answer:

work done =  665.12 kJ/k.mol  of nitrogen  

dQ = 685.905 kJ/k.mol of nitrogen

Explanation:

given data

pressure = 80 bar

temperature t1  = 220 K

temperature t2 =   300 K

solution

first we get here work done as considering ideal gas condition

work done = P (v2-v1 )   = nR (t2-t1)    

put here value

work done = 1 × 8.314 ×( 300 - 220 )

work done =  665.12 kJ/k.mol  of nitrogen  

and

now we get heat transfer by 1st law of thermodynamics that is

heat transfer dQ = dv + dw

dQ = Cv dT + dw

put here value and we get

dQ =  $\frac{R}{\gamma -1 }$   × (t2-t1) + 665.12

dQ = $\frac{8.314}{1.4-1}$  × (300-220) + 665.12

dQ = 685.905 kJ/k.mol of nitrogen