The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V = KT^(2/3). The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]

Solution

Physics sir · 4 steps

Write P in terms of T, write dV in terms of dT, and the work integral turns into a simple multiple of ΔT.

1. Pressure from T. Take 1 mole of gas. From PV = RT and V = KT^(2/3), the pressure is P = RT/KT^(2/3) = R/KT^(1/3).
2. Volume change. Differentiate V = KT^(2/3) to get dV = 2/3KT^(-1/3) dT.

   W = ∫ P dV = ∫ R/KT^(1/3) × 2/3KT^(-1/3) dT = 2/3R Δ T

3. Put in numbers. Both K and the powers of T cancel. With Δ T = 90 K, W = 2/3 × 90 R = 60R. The gas being monoatomic does not matter here, because work depends only on how P and V change.
4. So. W = 60R, so x = 60.

The common slip: 90. Treating the process as isobaric and using W = R Δ T gives 90, not 60. Here the pressure changes with T, so you must integrate.

The idea to keep. For any path given as V in terms of T, write P = RT/V and convert dV to dT. For PVᵐ = constant, the work is (R Δ T)/(1-m) per mole.

Answer from the NTA final answer key. Solution by PrepWiser.