A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This expansion

Question

A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This expansion is represented by the straight line from 1 to 2 in the PV diagram. The gas is then compressed isothermally. This compression is represented by the curved line from 2 to 1 in the PV diagram. Calculate the work per cycle done by the gas.

kJ

Please help. I already tried these answers and they were wrong :1.729 KJ and 6.6 KJ. Please show the correct answer with solution. I posted this question twice and no one got it correct. Thank you!

Explanation / Answer

Work done for expansion (from the diagram),W12 = 0.5*(24-12)*(2-1) + (24-12*1) atm.L <---- area of the traingle under the straigt line

So, W12 = 18 atm.L

Now, 1 atm = 1.013*10^5 Pa

1L = 10^-3 m3

So, W12 = 18*1.013*10^5*10^-3 = 1823.4 J

Workdone in isothermal compression is given by :

W = nRT*ln(Vf/Vi)

So, work done from 2 to 1, W21 = n*RT*ln(12/23)

where n = no of moles = 1

R = 0.0821 atm.L/mol.K

T = P*V/nR at point 2 = 1*24/(1*0.0821) = 292.3 K

So, W21 =1*0.0821*292.3*ln(12/24) atm.L = -16.63 atm.L

Now, 1 atm = 1.013*10^5 Pa

1L = 10^-3 m3

So, W21 = -16.63*1.013*10^5*10^-3 = -1684.6 J

So, net work done = W12 + W21 = 1823.4 - 1684.6 = 138.8 J = 0.139 kJ <-------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.