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 mole -1 K -1 )

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 mole -1 K -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

0.056 kg N 2=56 gm of N 2=2 mole of N 2 T 1=400 K , v α √ T so T 2=4 T 1=1600 K Q =( f /2) nR Δ T f =5 Q =12 k cal

Q. $0.056 k g$ of Nitrogen is enclosed in a vessel at a temperature of $12 7^{\circ} C$ . The amount of heat required to double the speed of its molecules is _____ $k$ cal. (Take $R = 2$ cal mole $^{- 1} K^{- 1}$ )

$0.056 k g N_{2} = 56 g m$ of $N_{2} = 2$ mole of $N_{2}$
$T_{1} = 400 K, vα T$ so $T_{2} = 4 T_{1} = 1600 K$
$Q = \frac{f}{2} n R Δ T$
$f = 5$
$Q = 12 k c a l$

Q. 0.056kg 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 mole −1K−1 )

0.056kgN2​=56gm of N2​=2 mole of N2​ T1​=400K,vαT​ so T2​=4T1​=1600K Q=2f​nRΔT f=5 Q=12kcal

Q. $0.056 k g$ of Nitrogen is enclosed in a vessel at a temperature of $12 7^{\circ} C$ . The amount of heat required to double the speed of its molecules is _____ $k$ cal. (Take $R = 2$ cal mole $^{- 1} K^{- 1}$ )

$0.056 k g N_{2} = 56 g m$ of $N_{2} = 2$ mole of $N_{2}$
$T_{1} = 400 K, vα T$ so $T_{2} = 4 T_{1} = 1600 K$
$Q = \frac{f}{2} n R Δ T$
$f = 5$
$Q = 12 k c a l$