An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature.
A. The work done by gas during the process is zero.
B. The heat added to gas is different from change in its internal energy.
C. The volume of the gas is increased.
D. The internal energy of the gas is increased.
E. The process is isochoric (constant volume process)
Choose the correct answer from the options given below:

Solution

Physics sir · 3 steps

Pressure rising in step with temperature means P/T stays fixed, so the volume never moves.

1. Read the graph. For an ideal gas PV = nRT, so V = nR/(P/T). A straight P–T line through the origin keeps P/T constant, so V is constant. The process is isochoric.

   P ∝ T ⇒ P/T = nR/V = constant ⇒ V = constant

2. Work and energy. With no change in volume, W = ∫ P dV = 0. Pressure rises, so temperature rises, and Δ U = nCᵥΔ T > 0. The first law then gives Q = Δ U + W = Δ U.

   Statement	Check	Verdict
   A: work done is zero	dV = 0, so W = 0	True
   B: Q differs from Δ U	Q = Δ U + 0	False
   C: volume increased	V stays constant	False
   D: internal energy increased	T rises, so Δ U > 0	True
   E: process is isochoric	P/T constant gives constant V	True

3. So. Statements A, D and E are correct, which is option (b).

The idea to keep. Turn any P–T or V–T statement into PV = nRT first. Once you know which variable is fixed, W, Δ U and Q follow from the first law.

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