
Air is filled in a bottle at atmospheric pressure and it is corked at \[35{}^\circ C\]. If the cork can come out at 3 atmospheric pressure, then up to what temperature should the bottle be heated in order to remove the cork.
\[A.\,325.5{}^\circ C\]
\[B.\,851{}^\circ C\]
\[C.\,651{}^\circ C\]
D. None of these
Answer
484.2k+ views
Hint: Gay-Lussac’s law, also called the pressure-temperature law is the concept of this question. Thus law states that, at a constant volume, the pressure is directly proportional to the temperature, that is, with an increase in the temperature, the pressure also increases and vice - versa. We will use this relation to solve this problem.
Formula used: \[\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}}\]
Complete step by step answer:
From given, we have the values of the pressure at two different cases and one of the values of the temperature is given.
The formula that relates the temperature and the pressure values at the constant volume is given by Gay-Lussac’s law, that is, as follows.
\[\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}}\]
Where \[{{T}_{1}},{{T}_{2}}\]are the temperatures values represented using the unit Kelvin and \[{{P}_{1}},{{P}_{2}}\] are the pressure values represented using the unit atm pressure.
At the temperature \[{{T}_{1}}=35{}^\circ C\] the pressure with which the air is filled in the bottle is equal to\[{{P}_{1}}=1\,atm\].
The unit of the temperature should be changed from Celsius to Kelvin. So, we have,
\[\begin{align}
& {{T}_{1}}=35{}^\circ C \\
& \Rightarrow {{T}_{1}}=35{}^\circ C+273 \\
& \Rightarrow {{T}_{1}}=308\,K \\
\end{align}\]
At the temperature \[{{T}_{2}}\] the pressure with which the air is filled in the bottle is equal to \[{{P}_{2}}=3\,atm\].
Substitute these values in the formula.
\[\begin{align}
& \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}} \\
& \Rightarrow\dfrac{{{T}_{2}}}{308}=\dfrac{3}{1} \\
\end{align}\]
Upon continuing the further calculation, we get the value of the temperature as,
\[\begin{align}
& {{T}_{2}}=3\times 308 \\
& \Rightarrow{{T}_{2}}=924\,K \\
\end{align}\]
As, in the options, the temperature is represented in terms of the unit Celsius, so, now we need to change the unit of the temperature to Kelvin.
\[\begin{align}
& {{T}_{2}}=924-273 \\
& \Rightarrow {{T}_{2}}=651\,{}^\circ C \\
\end{align}\]
So, the correct answer is “Option C”.
Note: The units of the parameters should be taken care of. The unit of the pressures given is the same, so, no need to change. The temperature is given in the degree Celsius, whereas, the law states that, the temperature in Kelvin increases or decreases at constant volume, along with the increase or decrease in the pressure. So, the unit of the parameter, the temperature should be changed from degree Celsius to Kelvin before solving the problem.
Formula used: \[\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}}\]
Complete step by step answer:
From given, we have the values of the pressure at two different cases and one of the values of the temperature is given.
The formula that relates the temperature and the pressure values at the constant volume is given by Gay-Lussac’s law, that is, as follows.
\[\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}}\]
Where \[{{T}_{1}},{{T}_{2}}\]are the temperatures values represented using the unit Kelvin and \[{{P}_{1}},{{P}_{2}}\] are the pressure values represented using the unit atm pressure.
At the temperature \[{{T}_{1}}=35{}^\circ C\] the pressure with which the air is filled in the bottle is equal to\[{{P}_{1}}=1\,atm\].
The unit of the temperature should be changed from Celsius to Kelvin. So, we have,
\[\begin{align}
& {{T}_{1}}=35{}^\circ C \\
& \Rightarrow {{T}_{1}}=35{}^\circ C+273 \\
& \Rightarrow {{T}_{1}}=308\,K \\
\end{align}\]
At the temperature \[{{T}_{2}}\] the pressure with which the air is filled in the bottle is equal to \[{{P}_{2}}=3\,atm\].
Substitute these values in the formula.
\[\begin{align}
& \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{P}_{1}}} \\
& \Rightarrow\dfrac{{{T}_{2}}}{308}=\dfrac{3}{1} \\
\end{align}\]
Upon continuing the further calculation, we get the value of the temperature as,
\[\begin{align}
& {{T}_{2}}=3\times 308 \\
& \Rightarrow{{T}_{2}}=924\,K \\
\end{align}\]
As, in the options, the temperature is represented in terms of the unit Celsius, so, now we need to change the unit of the temperature to Kelvin.
\[\begin{align}
& {{T}_{2}}=924-273 \\
& \Rightarrow {{T}_{2}}=651\,{}^\circ C \\
\end{align}\]
So, the correct answer is “Option C”.
Note: The units of the parameters should be taken care of. The unit of the pressures given is the same, so, no need to change. The temperature is given in the degree Celsius, whereas, the law states that, the temperature in Kelvin increases or decreases at constant volume, along with the increase or decrease in the pressure. So, the unit of the parameter, the temperature should be changed from degree Celsius to Kelvin before solving the problem.
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