A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by $Δ$ T, the weight becomes 150 gm. $Δ$ T should be :

27^{\circ}

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\frac{27^{\circ}}{4}

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300^{\circ}

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327^{\circ}

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Solution

The correct option is B $300^{\circ}$
Initially mass of air is 200 - 100 = 100gm. finally mass of air is 150 - 100 = 50 gm. As there is a hole in the wall, pressure inside the container will remain constant = $P_{0}$
$P V = n R T \Rightarrow T \propto \frac{1}{n}$
as number of moles of gas is halved, the temperature should be doubled (in K)
$T_{i} = 300 K$ So, $T_{f} = 600 K \Rightarrow Δ T = 300 K = 300^{\circ} C$

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A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by Δ T, the weight becomes 150 gm. Δ T should be :

A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by $Δ$ T, the weight becomes 150 gm. $Δ$ T should be :