Answer:
(a) The final pressure of the sample becomes one-fourth of the original pressure.
(b) The pressure of the sample remains unchanged.
(c) The final pressure of the sample becomes four times of the original pressure.
Explanation:
(a)
[tex]P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}[/tex]
The volume of sample doubled and kelvin temperature halved.
[tex]V_{2}=2V_{1}[/tex]
[tex]T_{2}=\frac{1}{2}T_{1}[/tex]
[tex]P_{2}=\frac{P_{1}\times V_{1}\times \frac{1}{2}T_{1}}{T_{1}\times2V_{1}}=\frac{P_{1}}{4}[/tex]
Therefore, the final pressure of the sample becomes one-fourth of the original pressure.
(b)
Volume and temperature of sample doubled.
[tex]V_{2}=2V_{1}[/tex]
[tex]T_{2}=2T_{1}[/tex]
[tex]P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times2V_{1}}={P_{1}}[/tex]
Therefore, the pressure of the sample unchanged.
(c)
Volume of sample halved and temperature double.
[tex]V_{2}=\frac{1}{2}V_{1}[/tex]
[tex]T_{2}=2T_{1}[/tex]
[tex]P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times \frac{1}{2}V_{1}}={4P_{1}}[/tex]
Therefore, the pressure of the sample becomes four times of the original pressure.
