A monoatomic gas performs a work of Q/4 where Q is the heat supplied to it. The molar heat capacity of the gas will be ______________ R during this transformation. Where R is the gas constant.

Solution

Physics sir · 4 steps

Work takes a quarter of the heat, so three quarters goes into internal energy. Use that to find C.

1. Split the heat. By the first law, Q = Δ U + W. With W = Q/4, the rest goes into internal energy: Δ U = 3Q/4.
2. Write both in ΔT. For n moles, Q = nC Δ T, where C is the molar heat capacity for this process. For any process, Δ U = nCᵥ Δ T, and a monoatomic gas has Cᵥ = 3/2R.

   n·3/2R Δ T = 3/4 nC Δ T

3. Solve for C. Cancel n Δ T to get 3/2R = 3/4C, so C = 4/3 × 3/2R = 2R.
4. So. The molar heat capacity during this transformation is 2R, so the answer is 2.

The common slip: 6. Taking Δ U = Q/4 instead of 3Q/4 gives C = 6R. The quarter is the work done, not the change in internal energy.

The idea to keep. When work is a fixed fraction of the heat, write Δ U = nCᵥΔ T and Q = nCΔ T. Then C = Cᵥ divided by the fraction of heat that goes into internal energy.

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