The temperature of a gas placed in an open container is raised from 27- to 227- C- The percent of the original amount of the gas expelled from the container will be -A- 20B- 40C- 60D- 80

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Avogadro's Law & Gay Lussac's Law

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Question

The temperature of a gas placed in an open container is raised from $27^{\circ} t o 227^{\circ} C .$ The percent of the original amount of the gas expelled from the container will be ?

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Solution

The correct option is B
40

During heating volume of container and pressure of the gas (= 1 atm) remain constant.
Hence, $n_{1} T_{1} = n_{2} T_{2}$
Or, $n_{2} = \frac{n_{1} T_{1}}{T_{2}} = n_{1} \times \frac{300}{500} = \frac{3}{5} n_{1}$ (moles remaining);
Moles expelled $= n_{1} - \frac{3}{5} n_{1} = \frac{2}{5} n_{1}$
% expelled $= \frac{2}{5} n_{1} \times \frac{100}{n_{1}} = 40$

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The temperature of a gas placed in an open container is raised from 27∘ to 227∘C. The percent of the original amount of the gas expelled from the container will be ?

The correct option is B 40 During heating volume of container and pressure of the gas (= 1 atm) remain constant. Hence, n1T1=n2T2 Or, n2=n1T1T2=n1×300500=35n1 (moles remaining); Moles expelled =n1−35n1=25n1 % expelled =25n1×100n1=40

The temperature of a gas placed in an open container is raised from $27^{\circ} t o 227^{\circ} C .$ The percent of the original amount of the gas expelled from the container will be ?