Helium gas at 104.0 kPa and 350 K is located within a cylinder with a piston. In

Question

Helium gas at 104.0 kPa and 350 K is located within a cylinder with a piston.  Initially the gas occupies 0.2900 m^3.  While a constant force, F, is applied to the end of the piston so that the pressure inside the cylinder is held constant at 104.0 kPa, 6110.0 J of heat is transferred to the helium gas.  The specific enthalpy of helium gas is given by the approximate relation:

     (H (kJ/mol) = 0.0208T(K))     

A) What is the final temperature of the helium gas?

B) What is the final volume of the helium gas?

C) What is the work done by the gas?

D) What is the change of internal energy of the system?

I appreciate any help you guys can give me!

Explanation / Answer

Cp = dH/dT = 0.0208 kJ/mol T = 20.8 J/mol T dQ = CpdT = 6110/n where of moles = PV/RT = 1.04 atm * 290L/0.082 * 350 = 10.5 moles

therefore dQ per mole = 6110/10.5 = 581.9 J/mol therefore change in temperature dT = 581.9/20.8 = 27.9 K

A)therefore final temperature of the gas = 377.9 K

B) pressure is constant therefore we use V1/T1 = V2/T2 therefore final volume V2 = 313.117 L

C) work done by gas = 104.0 * 10^3 N/m^2 * 0.313117 m^3 =32.564 kJ

D)change in internal energy = nCvdT n =10,5 Cv = Cp - R for monoatomic gas therefore since R =8.314 we get dU = 3658.94 J

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