0.056 kg of Nitrogen is enclose in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ___________ k cal.

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Vishal Baghel | Contributor-Level 10

6 months ago

$V_{r m s} α \sqrt[]{T} | T_{i} = 127 ° C = 400 k .$

$V_{r m s^{1}} = 2 V_{r m s}, m = 0.056 k g = 56 g$

T_f = 4T_i = 4 * 400 = 1600 k

$Q = n c_{v} Δ T = (\frac{m}{M}) C_{v} Δ T$

= $(\frac{56}{28}) \frac{5}{2} R Δ T$

= $2 * \frac{5}{2} * 2 * 1200$

Q = 12 * 10³ cal

= 12 k cal

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physics ncert solutions class 11th 2023

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