Question

1. One mole of an ideal monoatomic gas is initially at 300 K and 5 bar of pressure inside a cylinder with a frictionless piston.

   1. a) The cylinder is kept in a heat bath and the gas is allowed to expand under 1 bar

      of external pressure. Calculate the work and heat associated with this process.

   2. b) Calculate the change in enthalpy for isothermal expansion at constant pressure.

   3. c) Alternatively, the gas is allowed to expand isothermally under near-equilibrium

      conditions. Calculate the work and heat associated with this reversible process

Answer 1

a)

In this part of question volume change is not given and not mention process is isothermal so we can say that it is isochoric process. The work done by the system will be zero. Any heat energy transferred to the system will be absorbed as internal energy. ∆W= 0 and ∆U = ∆Q

b)

For an ideal gas the change in internal energy is zero (ΔU=0).

ΔH = ΔU+ ∆ (PV) = 0 + ∆(nRT) = 0

Change in enthalpy for isothermal expansion at constant pressure is zero.

c)

The cylinder be placed in a heat bath and the piston is moved outwards so the gas expands. During this process work some work done and gas is cool but it absorbs some heat from the source to keep it at the same temperature. Thus the expansion is isothermal.

Work done in isothermal process is ----------(1)

But in isothermal process T is constant so according to we can write

So we can write P₁V₁ = P₂V₂

                                                         = putting this value in equation (1)

We get ---------(2) Given value is n = 1, T = 300K, P₁ = 5 bar, P₂ = 1 bar R = 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹

Putting these values in equation (2)

                                     W = 5744.14× 0.69897 = 4014.98 joul

The change in internal energy of gas is 0 joules in an isothermal process. So Q = W, the heat added to the gas is also equal to 4014.98 joules.