Calculate the amount of work done in each of the following cases : (i

1.	Calculate the amount of work done in each of the following cases : (i) One mole of an ideal gas contained in a bulb of 10 litre capacity at 1 bar is allowed to enter into an evacuated bulb of 100 litre capacity. (ii) One mole of a gas is allowed to expand from a volume of 1 litre to a volume of 5 litres against the constant external pressure of 1 atm (1 litres atm = 101.3 J) Calculate the internal energy change (ΔU) in each case if the process were carried out adiabatically.
Answer» (i) w = –P_ext × ΔV As expansion taks place into the evacuated bulb, i.e., against vacuum, P_ext = 0. Henc, w = 0. For adiabatic process, q = 0 ∴ ΔU = q + w = 0 + 0 = 0. (ii) V = V₂ – V₁ = 5 – 1 = 4 litres P = 1 atm ∴ w = – PV = –1 × 4 litre atm = – 4 litres atm = – 4 × 101.3 J = – 405.2 J (1 L – atm = 101.3J) The negative sign implies that the work is done by the system. For adiabatic process, ΔU = q + w = 0 – 405.2 J = – 405.2 J.

Calculate the amount of work done in each of the following cases : (i) One mole of an ideal gas contained in a bulb of 10 litre capacity at 1 bar is allowed to enter into an evacuated bulb of 100 litre capacity. (ii) One mole of a gas is allowed to expand from a volume of 1 litre to a volume of 5 litres against the constant external pressure of 1 atm (1 litres atm = 101.3 J) Calculate the internal energy change (ΔU) in each case if the process were carried out adiabatically.

Answer»

(i) w = –P_ext × ΔV

As expansion taks place into the evacuated bulb, i.e.,

against vacuum, P_ext = 0. Henc, w = 0.

For adiabatic process, q = 0

∴ ΔU = q + w = 0 + 0 = 0.

(ii) V = V₂ – V₁ = 5 – 1 = 4 litres

P = 1 atm ∴ w = – PV

= –1 × 4 litre atm = – 4 litres atm

= – 4 × 101.3 J = – 405.2 J (1 L – atm = 101.3J)

The negative sign implies that the work is done by the system.

For adiabatic process, ΔU = q + w = 0 – 405.2 J = – 405.2 J.