A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2° C. The work done by the gas is: Given, R=8.3 J K⁻¹ mol⁻¹.

Solution

Physics sir · 4 steps

Helium is monatomic, so find Δ U from Cᵥ = 3/2R; the rest of the heat is the work.

1. First law. The heat given splits into two parts: raising the internal energy and doing work. So Q = Δ U + W, which gives W = Q - Δ U.
2. Internal energy change. Helium is monatomic, so Cᵥ = 3/2R. A rise of 2°C is the same as a rise of 2 K.

   Δ U = nCᵥΔ T = 1 × 3/2 × 8.3 × 2 = 24.9 J

3. Work done. W = Q - Δ U = 48 - 24.9 = 23.1 J.
4. So. The gas does 23.1 J of work, which is option (a).

The idea to keep. Internal energy of an ideal gas always changes by nCᵥΔ T, whatever the process. Find Δ U that way, then the first law gives you the work.

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