Two flasks A and B have equal volumes. A is m | vedclass

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(B) The total kinetic energy $(K.E.)$ of an ideal gas is given by the formula: $K.E. = \frac{3}{2} nRT$. For flask $A$ containing $H_2$: $n_A = \frac{

Vedclass · Accepted Answer

$11:1$

Vedclass · Answer

(B) The total kinetic energy $(K.E.)$ of an ideal gas is given by the formula: $K.E. = \frac{3}{2} nRT$. For flask $A$ containing $H_2$: $n_A = \frac{m}{M_{H_2}} = \frac{m}{2}$ and $T_A = 300 \ K$. Thus,$(K.E.)_A = \frac{3}{2} 	imes \frac{m}{2} 	imes R 	imes 300$. For flask $B$ containing $CO_2$: $n_B = \frac{m}{M_{CO_2}} = \frac{m}{44}$ and $T_B = 600 \ K$. Thus,$(K.E.)_B = \frac{3}{2} 	imes \frac{m}{44} 	imes R 	imes 600$. Taking the ratio: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{\frac{m}{2} 	imes 300}{\frac{m}{44} 	imes 600} = \frac{300/2}{600/44} = \frac{150}{600/44} = \frac{150 	imes 44}{600} = \frac{44}{4} = 11:1$.

Two flasks $A$ and $B$ have equal volumes. $A$ is maintained at $300 \ K$ and $B$ at $600 \ K$. Equal masses of $H_2$ and $CO_2$ are taken in flasks $A$ and $B$ respectively. Find the ratio of total $K.E.$ of gases in flask $A$ to that of $B$.

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