JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 19)
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)
Answer
12
Explanation
Because the vessel is closed, it will be an isochoric process.
To double the speed, temperature must be 4 times (v $$\alpha$$$$\sqrt{T}$$)
So, Tf = 1600 K, Ti = 400 K
number of moles are $${{56} \over {28}} = 2$$
so Q = nCv $$\Delta$$T = 2 $$\times$$ $${5 \over 2}$$ $$\times$$ 2 $$\times$$ 1200
= 12000 cal = 12 K cal
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