
A gas of mass \[{m_o}\] is filled in a vessel at \[7^\circ C\]. If \[x\] fraction of its mass escapes out at \[27^\circ C\], find \[x\]. (Assuming pressure constant).
Answer
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Hint: The gas escapes at constant pressure means that the gas follows Charles’s law. The temperature and volume changes when gas expands.
Complete step by step answer:
Let the volume of the gas at \[7^\circ C\] = \[{V_1}\]
The volume of the gas at \[27^\circ C\] = \[{V_2}\]
And \[{V_2} = {V_1} - x\]
Following Charles’s law,
\[\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}\]
Given \[{T_1} = 7^\circ C = 273 + 7 = 280K\],
\[{T_2} = 27^\circ C = 273 + 27 = 300K\]
Inserting in the equation,
$\dfrac{{{V_1}}}{{280}} = \dfrac{{{V_2}}}{{300}}$
$\dfrac{{{V_1}}}{{280}} = \dfrac{{{V_1} - x}}{{300}}$
$300{V_1} = 280{V_1} - 280x$
$280x = - 20{V_1}$
$x = - \dfrac{1}{{14}}{V_1}$
The\[x\] is the $\dfrac{1}{{14}}$ times the volume of the gas filled at \[7^\circ C\].
Additional Information:
Charles’s law demonstrates the way in which the gas tends to expand when heated. In exact words, keeping pressure to be constant the volume of the gas is directly proportional to absolute temperature.
The relationship is shown as
$V \propto T$
$\dfrac{V}{T} = k$, \[k\] is a non zero constant
From the above relation it is said that if the temperature of the gas increases that volume of the gas increases and vice versa.
Unlike Charles’s law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature stated by Boyle’s law.
$P \propto \dfrac{1}{V}$
$PV = k$, \[k\] is non-zero constant.
Another law states that the pressure of the gas varies directly to the temperature of the gas at constant volume.
$P \propto T$
$\dfrac{P}{T} = k$, \[k\] is non zero constant
Note:
The combined form of gas laws are given by the relation $\dfrac{{PV}}{T} = k$. The physical changes in the surroundings are all various forms of gas laws. For example, in summer the pressure is also high when temperature is more, so the gas tires get burst often. Inhalation problems also happen during the climb of mountains with decrease in pressure as the volume of oxygen decreases.
Complete step by step answer:
Let the volume of the gas at \[7^\circ C\] = \[{V_1}\]
The volume of the gas at \[27^\circ C\] = \[{V_2}\]
And \[{V_2} = {V_1} - x\]
Following Charles’s law,
\[\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}\]
Given \[{T_1} = 7^\circ C = 273 + 7 = 280K\],
\[{T_2} = 27^\circ C = 273 + 27 = 300K\]
Inserting in the equation,
$\dfrac{{{V_1}}}{{280}} = \dfrac{{{V_2}}}{{300}}$
$\dfrac{{{V_1}}}{{280}} = \dfrac{{{V_1} - x}}{{300}}$
$300{V_1} = 280{V_1} - 280x$
$280x = - 20{V_1}$
$x = - \dfrac{1}{{14}}{V_1}$
The\[x\] is the $\dfrac{1}{{14}}$ times the volume of the gas filled at \[7^\circ C\].
Additional Information:
Charles’s law demonstrates the way in which the gas tends to expand when heated. In exact words, keeping pressure to be constant the volume of the gas is directly proportional to absolute temperature.
The relationship is shown as
$V \propto T$
$\dfrac{V}{T} = k$, \[k\] is a non zero constant
From the above relation it is said that if the temperature of the gas increases that volume of the gas increases and vice versa.
Unlike Charles’s law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature stated by Boyle’s law.
$P \propto \dfrac{1}{V}$
$PV = k$, \[k\] is non-zero constant.
Another law states that the pressure of the gas varies directly to the temperature of the gas at constant volume.
$P \propto T$
$\dfrac{P}{T} = k$, \[k\] is non zero constant
Note:
The combined form of gas laws are given by the relation $\dfrac{{PV}}{T} = k$. The physical changes in the surroundings are all various forms of gas laws. For example, in summer the pressure is also high when temperature is more, so the gas tires get burst often. Inhalation problems also happen during the climb of mountains with decrease in pressure as the volume of oxygen decreases.
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