Chapter3, Lesson A - Enthalpy and Internal Energy for Ideal Gases

3A-1 :	Enthalpy and Internal Energy for Ideal Gases							2 pts

Determine the specific internal energy of 98 L of neon gas at 400°C and 2 MPa. The neon gas has a total enthalpy of 1200 kJ . Assume that the neon gas is an ideal gas.

Read :	Given the temperature, pressure and volume of neon in an ideal gas state, we can calculate the mass of neon in the system using the Ideal Gas EOS. This allows us to convert the enthalpy into specific enthalpy. We can use the definition of enthalpy or specific enthalpy to relate U to H and PV and then eliminate PV using the Ideal Gas EOS again. The units may get tricky.




Given:	V =	98	L		P =	2	MPa
	T =	400	oC		H =	1200	kJ

Find:	U =	???	kJ/kg

Assumptions:		- Equilibrium conditions
		- Neon is an ideal gas at this T and P

Solution :	Since neon behaves as an ideal gas, the definition of specific enthalpy can be modified as follows:

		Eqn 1		Eqn 2				Eqn 3

	But :		Eqn 4					Eqn 5

	For ideal gases :			Eqn 6				Eqn 7

	Molecular weight of neon : ( NIST WebBook )			MW	20.18	g / mol
	Universal Gas Constant values :			R	0.08205	atm L/gmol K
				R	8.314	J/mol K or Pa m3/mol K

	Note: To convert °C to K, add 273.15 to °C.				T =	673.15	K

					m =	0.7067	kg
					RT / MW	277.3	kJ/kg
					H =	1698.0	kJ/kg

					U =	1421	kJ/kg

Example Problem with Complete Solution