Assignment 10.1: An ideal gas absorbs 5.00 × 103 J of energy while doing 2.00 × 103 J of work on the environment during a constant pressure process. (a) Compute the change in the internal energy of gas. (b) If the internal energy now drops by 4.50 × 103 J and 7.50 × 103 J is expelled from the system find the change in volume assuming a constant pressure process at 1.01 × 105 Pa.
Solution (a): Given that heat absorbed, ΔQ = 5.00 × 103 J and work done = 2.00 × 103 J, apply First Law of Thermodynamics, ΔQ = Δu + ΔW ⇒ 5.00 × 103 = Δu + 2.00 × 103 ⇒ 5.00 × 103 – 2.00 × 103 = Δu ⇒ Δu = 3 × 10-3 J.
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Solution (b): Given: Constant pressure, P = 1.01 × 105 Pa
Δu = 4.50 × 103 J, ΔQ = 7.50 × 103 J
Required: Change in volume, ΔV
Formulae: ΔQ = Δu + ΔW … (1)
For constant pressure process, ΔW = PΔV … (2)
Calculation: Put value of ΔW in the first formula, ΔQ = Δu + PΔV. Now put given values in this equation,
Since heat is expelled, internal energy drops and pressure is constant, therefore, there would be a decrease in the volume. So the change in volume is negative.