The document describes a gas confined in a container that is heated, causing it to expand against a constant external pressure. It provides the initial conditions and asks to calculate: (1) The new volume after expansion is 0.4 L (2) The work done during expansion is 0.1974 l-atm (3) The change in internal energy is 0 since temperature is constant (4) The change in enthalpy is 2 J, equal to the heat supplied

1. (c) 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: (i) the new volume it occupies after the expansion (i) the work done in
the expansion (iii) the change in internal energy of the gas (iv) the change in enthalpy of the gas.
(4 marks) (2 marks) (2 marks) (3 marks)
Solution
c) initial volume(v1) = 0.2 L = 200 cm3
external pressure(P) = 1 bar = 0.987 atm
heat supplied to system = 2 j
we know that, 1 l.atm = 101.3 joule
amount of work can be done with supplied energy = 2/101.3 = 0.1974 l.atm
i) w = - P(V2-V1)
V1 = 0.2 l V2 = final volume = ?
P = 0.987 atm
-0.1974 = 0.987*(V2-0.2)
V2 = 0.4 L
ii) work done in the expansion = 0.1974 l.atm
iii) as the temperature is constant
change in internal energy(DU) = 0
iv) DH = DU + PDV

2. = 0 + 0.1974*101.3
= 2j