A vessel contains 14 g of nitrogen gas at a temperature of 27° C. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be: (Take R =8.32 J mol -1 k -1 )

Question

A vessel contains 14 g of nitrogen gas at a temperature of 27° C. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be: (Take R =8.32 J mol -1 k -1 ) (A) 2229 J (B) 5616 J (C) 9360 J (D) 13,104 J

Tardigrade Admin · Accepted Answer

v rms propto √ T v rms propto √300 K , v rms f =2 v rms i v rms f propto √1200 K T f =1200 K , T i =300 K , n =(14/28)=(1/2) Q = nC v Δ T =(1/2) × (5 R /2) × 900 Q =9360 J

Q. A vessel contains $14 g$ of nitrogen gas at a temperature of $2 7^{\circ} C$ . The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be : (Take $R = 8.32 J m o l^{- 1} k^{- 1}$ )

$2229 J$

$5616 J$

$9360 J$

$13, 104 J$

$v_{r m s} \propto T$
$v_{r m s} \propto 300 K, v_{r m s_{f}} = 2 v_{r m s_{i}}$
$v_{r m s_{f}} \propto 1200 K$
$T_{f} = 1200 K, T_{i} = 300 K, n = \frac{14}{28} = \frac{1}{2}$
$Q = n C_{v} Δ T = \frac{1}{2} \times \frac{5 R}{2} \times 900$
$Q = 9360 J$

Q. A vessel contains 14g of nitrogen gas at a temperature of 27∘C. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be : (Take R=8.32Jmol−1k−1 )

2229J

5616J

9360J

13,104J