JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Heat & Thermodynamics

A vessel contains 14 g of nitrogen gas at a temperature of

2 7^{\circ} C

. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be : (Take

R = 8.32 J mol^{- 1} k^{- 1}

)

(A)2229 J
(B)5616 J
(C)9360 J✓ Correct
(D) $13, 104 J$

Answer: (C)

Solution

Since

v_{r m s} \propto T

, doubling

v_{r m s}

requires quadrupling T. Initial temperature

T_{i} = 300

K, so final temperature

T_{f} = 1200

K. Number of moles

n = 14/28 = 0.5

. For diatomic nitrogen,

C_{v} = (5/2) R

. Heat required:

Q = n C_{v} Δ T = 0.5 \times (5/2) \times 8.32 \times 900 = 9360

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option

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