When $0.2$ mole of ethane is burnt completely in a bomb calorimeter, the temperature of calorimeter system increased by $2 K$ . What should be the increase in temperature of the same calorimeter system, when $0.4$ mole of ${C H}_{4}$ is burnt?
Given: $Δ_{C} U_{C_{2} H_{6} (g)} = - 400 K c a l / m o l$ and $Δ_{c} U_{C H_{4} (g)} = - 200 K c a l / m o l$ .

2 K

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4 K

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1 K

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3 K

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Solution

The correct option is A $2 K$
$1 m o l$ of $C H_{4}$ produces $200 K$ calories
While $1 m o l$ of $C_{2} H_{6}$ produce $400 K$ calories
So, $0.2 m o l C_{2} H_{6}$ produce $0.2 \times 400 K c a l$
$= 80 K c a l$
$0.4 m o l C H_{4}$ produce $= 0.4 \times 200$
$= 80 K c a l$
So the raise in temperature is same for both $0.2 m o l$ ethane and $0.4 m o l$ methane $= 2$ Kelvin

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