Question

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ___ k cal.(Take R = 2 cal mol–1 K–1)

Answer 1

Because the vessel is closed, it will be an isochoric process.
To double the speed, the temperature must be 4 times
(v∝$\sqrt T$)
So T f = 1600 K, T i = 400 K
number of moles are [$\frac{56}{28}=2$]
So, Q=nCvΔT=2×$\frac{5}{2}$×2×1200
=12000 cal
=12 K cal