10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P₁ to P₂ is α Joule ( P₁=21.7 Pa and P₂=30 Pa,Cᵥ=21 J/K.mol,R=8.3 J/mol.K). The value of α is ____ .

Solution

Physics sir · 3 steps

P1 to P2 is a vertical line, so volume is fixed: no work is done, and all the heat goes into internal energy.

1. Read the graph. The path from P₁ to P₂ is vertical at V = 1 m³. The process is isochoric, so W = 0 and Q = Δ U = nCᵥ Δ T.
2. Find the temperature change. From PV = nRT at constant V, nR Δ T = V Δ P. Here Δ P = 30 - 21.7 = 8.3 Pa, so Δ T = (1 × 8.3)/(10 × 8.3) = 0.1 K.

   Q = nCᵥ Δ T = 10 × 21 × 0.1 = 21 J

3. So. The heat absorbed from P₁ to P₂ is α = 21 J, option (a).

The idea to keep. A vertical line on a P-V graph means constant volume. Use Q = nCᵥΔ T, and get nRΔ T straight from VΔ P without finding each temperature.

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