Answer:
The volume of the gas decreases 140 times.
Explanation:
The equation of state of the ideal gas is described below:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure.
[tex]V[/tex] - Volume.
[tex]n[/tex] - Molar quantity.
[tex]R_{u}[/tex] - Ideal gas constant.
[tex]T[/tex] - Temperature.
From this formula we construct the following relationship:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex] (2)
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressure.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volume.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperature.
[tex]\frac{V_{2}}{V_{1}} = \frac{\frac{T_{2}}{T_{1}} }{\frac{P_{2}}{P_{1}} }[/tex]
If we know that [tex]\frac{T_{2}}{T_{1}} = \frac{1}{14}[/tex] and [tex]\frac{P_{2}}{P_{1}} = 10[/tex], then the change in the volume of the gas is:
[tex]\frac{V_{2}}{V_{1}} = \frac{\frac{1}{14} }{10}[/tex]
[tex]\frac{V_{2}}{V_{1}} = \frac{1}{140}[/tex]
The volume of the gas decreases 140 times.
