
An open flask containing air is heated from $ {\text{300K}} $ to $ {\text{500K}} $ .What is the percentage of air escaped to the atmosphere?
(A) 20
(B) 40
(C) 60
(D) 80
Answer
561k+ views
Hint: In the above question, it is asked about the percentage of air escaped. So first, we will find the volume of air present when temperature is increased using Charles law. Then, after getting the volume to be present we can find the air escaped and hence, we can find the percentage of air escaped.
Complete step by step solution
Charles' law described how a gas behaves when it gets heated. The gas usually gets expanded. So we can say that according to the Charles law, volume and temperature are directly proportional to each other when pressure applied to the system is constant, i.e.,
$ {\text{V}} \propto {\text{T}} $
which implies that volume to the temperature ratio is always constant.
$ \dfrac{{\text{V}}}{{\text{T}}}{\text{ = constant}} $
Hence, we can say that the $ \dfrac{{\text{V}}}{{\text{T}}} $ ratio at temperature $ T_1 $ is equal to $ \dfrac{{\text{V}}}{{\text{T}}} $ ratio at temperature $ T_2 $ .
$ \dfrac{{{{V_1}}}}{{{{T_1}}}}{\text{ = }}\dfrac{{{{V_2}}}}{{{{T_2}}}} $
Let $ {{V_1 = V}} $
Rearranging the above equations, we get:
$ {{V_1 = }}\dfrac{{{{V_1 \times T_2}}}}{{{{T_1}}}} $
It is given that $ {{T_1 = 300K}} $ and $ {{T_2 = 500K}} $ . So, after substituting the values of $T_2$,$ V_1$ and $ T_1$ , we get:
$ {{V_1 = }}\dfrac{{{{V \times 500}}}}{{300}} $
$ {{V_1 = }}\dfrac{{{{5V}}}}{3} $
Now, volume of air escaped ( $ {{\text{V}}_{\text{e}}} $ ) = \[{V_2} - {V_1}\]= $ \dfrac{{{\text{5V}}}}{{\text{3}}}{\text{ - V}} $ =\[\dfrac{{{\text{5V - 3V}}}}{{\text{3}}}{\text{ = }}\dfrac{{{\text{2V}}}}{{\text{3}}}\]
Percentage of volume of air escaped = $ \dfrac{{{{{V}}_{{e}}}}}{{{{V_1}}}}{{ \times 100}} $ = $ \dfrac{{\dfrac{{{\text{2V}}}}{{\text{3}}}}}{{\dfrac{{{\text{5V}}}}{{\text{3}}}}}{{ \times 100 = }}\dfrac{{\text{2}}}{{\text{5}}}{{ \times 100 = 40}} $
$ \therefore $ The percentage of volume of air escaped is $ 40\% $ .
So, the correct option is option B.
Note
In these types of questions where there is a question to find out volume when temperature is given or vice-versa, we should apply Charles Law.
Charles Law is only applicable to an ideal gas. We assume, air given in the question is ideal. Also, Charles Law is not applicable at low temperature.
Complete step by step solution
Charles' law described how a gas behaves when it gets heated. The gas usually gets expanded. So we can say that according to the Charles law, volume and temperature are directly proportional to each other when pressure applied to the system is constant, i.e.,
$ {\text{V}} \propto {\text{T}} $
which implies that volume to the temperature ratio is always constant.
$ \dfrac{{\text{V}}}{{\text{T}}}{\text{ = constant}} $
Hence, we can say that the $ \dfrac{{\text{V}}}{{\text{T}}} $ ratio at temperature $ T_1 $ is equal to $ \dfrac{{\text{V}}}{{\text{T}}} $ ratio at temperature $ T_2 $ .
$ \dfrac{{{{V_1}}}}{{{{T_1}}}}{\text{ = }}\dfrac{{{{V_2}}}}{{{{T_2}}}} $
Let $ {{V_1 = V}} $
Rearranging the above equations, we get:
$ {{V_1 = }}\dfrac{{{{V_1 \times T_2}}}}{{{{T_1}}}} $
It is given that $ {{T_1 = 300K}} $ and $ {{T_2 = 500K}} $ . So, after substituting the values of $T_2$,$ V_1$ and $ T_1$ , we get:
$ {{V_1 = }}\dfrac{{{{V \times 500}}}}{{300}} $
$ {{V_1 = }}\dfrac{{{{5V}}}}{3} $
Now, volume of air escaped ( $ {{\text{V}}_{\text{e}}} $ ) = \[{V_2} - {V_1}\]= $ \dfrac{{{\text{5V}}}}{{\text{3}}}{\text{ - V}} $ =\[\dfrac{{{\text{5V - 3V}}}}{{\text{3}}}{\text{ = }}\dfrac{{{\text{2V}}}}{{\text{3}}}\]
Percentage of volume of air escaped = $ \dfrac{{{{{V}}_{{e}}}}}{{{{V_1}}}}{{ \times 100}} $ = $ \dfrac{{\dfrac{{{\text{2V}}}}{{\text{3}}}}}{{\dfrac{{{\text{5V}}}}{{\text{3}}}}}{{ \times 100 = }}\dfrac{{\text{2}}}{{\text{5}}}{{ \times 100 = 40}} $
$ \therefore $ The percentage of volume of air escaped is $ 40\% $ .
So, the correct option is option B.
Note
In these types of questions where there is a question to find out volume when temperature is given or vice-versa, we should apply Charles Law.
Charles Law is only applicable to an ideal gas. We assume, air given in the question is ideal. Also, Charles Law is not applicable at low temperature.
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