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 )

true

explanation:

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.