0.056 kg of Nitrogen is enclosed in a vessel at a temperature of $127^{\circ} \mathrm{C}$. The amount of heat required to double the speed of its molecules is ______ k cal. (Take $\mathrm{R}=2$ cal mole$^{-1} \mathrm{~K}^{-1}$ )

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Heat & Thermodynamics

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of

12 7^{\circ} C

. The amount of heat required to double the speed of its molecules is ______ k cal. (Take

R = 2

cal mole

^{- 1} K^{- 1}

)

Answer: 12

Solution

Mass of

N_{2} = 0.056

= 56

= 2

moles (molecular mass = 28 g/mol).

Initial temperature

T_{1} = 127°

= 400

K.

Since rms speed

v \propto T

, to double the speed we need

T_{2} = 4 T_{1} = 1600

K.

For a diatomic gas,

f = 5

degrees of freedom.

Heat required:

Q = \frac{f}{2} n R Δ T = \frac{5}{2} \times 2 \times 2 \times (1600 - 400)

Q = 5 \times 2 \times 1200 = 12000

cal

= 12

kcal

The answer is 12.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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