A 2.5 mole sample of carbon dioxide gas (CO2) is initially at a temperature of 2

Question

A 2.5 mole sample of carbon dioxide gas (CO2) is initially at a temperature of 20°C. The pressure of the gas is held constant while the gas is heated to a temperature of 160°C.

a) What does equal for carbon dioxide gas?

b) How much heat is added to the gas during the process?

c) What is the change in internal energy of the gas during the process?

d) How much work is done by the gas during the process?

e) What would be the answers to parts b), c) and d) if the volume of the gas was held constant (instead of the pressure) as the gas was heated from 20°C to 160°C?

Explanation / Answer

a)

gamma r = 1.28

b)

Cp = gamma*Cv = gamma*(Cp-R)

Cp*( gamma - 1 ) = gamma*R

Cp = gamma*R/(gamma-1)

dQ = n*Cp*dT = n*gamma*R*dT/(gamma-1)

dQ = (2.5*1.28*8.314*(160-20))/(1.28-1)

dQ = 13302.4 J

(c)

Cv = cP/gamma = R/(gamma-1)

dU = n*Cv*dT = n*R*dT/(gamma-1) = (2.5*8.314*(160-20))/(1.28-1) = 10392.5 J

d)

W = dQ - dU = 2909.9 J

e)

dQ = n*Cv*dT = n*R*dT/(gamma-1) = (2.5*8.314*(160-20))/(1.28-1) = 10392.5 J

dU = n*Cv*dT = n*R*dT/(gamma-1) = (2.5*8.314*(160-20))/(1.28-1) = 10392.5 J

W = dQ - dw = 0

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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