1 text{ mole} of gas A and 1 text{ mole} of g | vedclass

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(B) The kinetic energy for $n$ moles of an ideal gas is given by $K.E. = \frac{3}{2} pV$. Given: $K.E. = 98.03 \ L \ atm$ and $V = 24.6 \ L$. Substitu

Vedclass · Accepted Answer

$2.66$

Vedclass · Answer

(B) The kinetic energy for $n$ moles of an ideal gas is given by $K.E. = \frac{3}{2} pV$. Given: $K.E. = 98.03 \ L \ atm$ and $V = 24.6 \ L$. Substituting the values: $98.03 = \frac{3}{2} 	imes p 	imes 24.6$. $p = \frac{98.03 	imes 2}{3 	imes 24.6} = \frac{196.06}{73.8} \approx 2.66 \ atm$. Since the reaction $A_{(g)} + B_{(g)} \longrightarrow 2 D_{(g)}$ involves $2 	ext{ moles}$ of reactants producing $2 	ext{ moles}$ of products, the total number of moles remains constant ($2 	ext{ moles}$ total). Thus, the pressure exerted by the gas $D$ at the end of the reaction is $2.66 \ atm$.

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