Verified Solution • 2025–26 Session moderate

QJEEM22Ph4ed7: JEE Main 2022 Physics - Class 11 JEEM22Ph4ed7 Solution: 0.056 kg of Nitrogen is enclosed in a vessel at a temperature…

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127 ∘^\circ C. Th amount of heat required to double the speed of its molecules is ____________ k cal.

Take R = 2 cal mole −- 1 K −- 1)

Step-by-Step Verified Solution

Because the vessel is closed, it will be an isochoric process.

To double the speed, temperature must be 4 times (v α\alpha T\sqrt{T} )

So, Tf = 1600 K, Ti = 400 K

number of moles are 5628=2{{56} \over {28}} = 2

so Q = nCv Δ\Delta T = 2 ×\times 52{5 \over 2} ×\times 2 ×\times 1200

= 12000 cal = 12 K cal

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