JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Chemical Thermodynamics

WhatsApp Email

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are:

Given:

R = 8.3 J K^{- 1} mol^{- 1}

(A)348 K and 300 J✓ Correct
(B)378 K and 300 J
(C)368 K and 500 J
(D)378 K and 500 J

Answer: (A)

Solution

Argon is monatomic, so

C_{p} = \frac{5}{2} R

and

C_{v} = \frac{3}{2} R

. Heat is supplied at constant pressure (1 atm), so

q = n C_{p} Δ T

500 = 0.5 \times \frac{5}{2} \times 8.3 \times Δ T \Rightarrow Δ T \approx 48 K

, giving

T_{f} \approx 346 - 348 K

. The internal-energy change

Δ U = n C_{v} Δ T = q \frac{C _{v}}{C _{p}} = 500 \times \frac{3}{5} = 300 J

. Hence

T_{f} \approx 348 K

and

Δ U = 300 J

(option A). NTA's 500 J would require a constant-volume process, contradicting the stated 1 atm.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Chemical Thermodynamics →

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Chemical Thermodynamics

Solution

More JEE Main questions on Chemical Thermodynamics