Question

A piston containing 2.00 mol of Ne at 298 K expands isothermally and reversiblyto double its volume.

a) Calculate q,w and deltaU

b) How would the values of q, w, and deltaU change (qualitatively) if this piston's expansion is irreversible?

I need part b please thank you

Answer 1

b. For both processes dU will be zero, as the processes are isothermal.

dU= dq + dw = 0

dw=-dq

For reversible changes the system is restored to its original state by traversing the forward sequence of steps in the reverse order, not only the system but also the surroundings are restored to their original states. A reversible expansion process can be imagined as expansion in infinite stages, where p_external = p_internal - dp; magnitude of dp is infinitesimally small. The work done by the system,

w = -p_external .dv = - ^v2∫_v1 p_external.dv = - ^v2∫_v1 (p_internal - dp).dv = - ^v2∫_v1 p_internal.dv .....[as dp is very small]

p_internal = nRT/V

w = - ^v2∫_v1 (nRT/V).dv = - nRT ln V₂/V₁

In case of irreversible processes you cannot write, p_external = p_internal

here, w = - p_exteral (V₂-V₁)

You need to know the external pressure to calculate the work done for the irreversible process.

The magnitude of the work involved in a reversible process is larger than the work involved in an irreversible expansion. We can prove this.

│w_rev │= nRT ln V₂/V₁ = nRTln[1+( V₂/V₁ -1)]

= nRT {( V₂/V₁ -1) + higher terms} .....[expanding the logarithm terms]

= (nRT/V₁)(V₂-V₁) + higher terms

= p_internal(V₂-V₁) + higher terms

and │w_irr│= p_exteral (V₂-V₁)

│w_rev│-│w_irr│= p_internal(V₂-V₁) + higher terms - p_exteral (V₂-V₁)

= (V₂-V₁)(p_internal-p_external) + higher terms

in expansion , V₂>V₁ and p_internal>p_external, so

│w_rev│-│w_irr│= always positive