Question

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.

Answer 1

Answer:

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 + W.....first 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 = - ∫ PdV.......work 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 + PV.....ideal 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 )