In a 200 cm3 container, a certain amount of o2 gas is held with a piston at equilibrium with the external pressure being 1 bar. the temperature of the gas is 298 k. when 2 j of heat is supplied, the gas expands against the constant external pressure. assuming that no heat is lost from the gas, calculate: (c) (i) the new volume it occupies after the expansion (4 marks) (2 marks) (2 marks) (3 marks) (ii) the work done in the expansion (ii) the change in internal energy of the gas (iv) the change in enthalpy of the gas.

a) V2 = 200.161 cm³

b)  Wexp = 0.0161 J done by the system

c) ΔU = 1.984 J

d) ΔH = 2.006 J

Explanation:

ΔU = Q + Wfirst law

a) at constant temperature (298 K ):

∴ ΔU = 0 ⇒ Q = - W

∴ W = - ∫ P dV

⇒ Q = ∫ PdV

∴ PV = nRT

⇒ P = nRT / V

⇒  Q = ∫ nRT dV/V

⇒ Q = nRT Ln ( V2 / V1 )assuming n O2 = 1 mol

⇒  Q / nRT = Ln ( V2 / V1 )

⇒ 2 J / (( 1mol) ( 8.314 J/molK * 298 K )) = Ln ( V2 / V1 )

⇒ 8.072 E-4 = Ln ( V2 / V1 )

⇒ 1.001 = V2 / V1

⇒ V2 = 1.001 * 200 cm³

⇒ V2 = 200.161 cm³  ( 2.00161 E-4 m³)

b) Wexp = - ∫ PdVwork done by the system ( - )

P = 1 bar = 10197.2 Kgf/m²

⇒ W exp = - 10197.2 Kgf/m² * ( 2.00161 E-4 - 2.00 E-4 )m³

⇒ W exp = - 1.642 E-3 Kgf *m ( - 0.0161 J )

c) ΔU = Q + W

⇒ ΔU = 2 J - 0,0161 J

⇒ ΔU = 1.984 J   ( 0.202 Kgf*m )

d) H = U + PVideal gas

⇒ ΔH = Δ ( U + PV )

⇒ ΔH = ΔU + P2V2 - P1V1

⇒ ΔH = 0.202  + (( 10197.2 * 2.00161 e-4 )) - (( 10197.2 * 2.00 E-4 ))

⇒ ΔH = 0.202 + 2.041 - 2.0394

⇒ ΔH = 0,204 Kgf*m ( 2.006 J )

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .