Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.
Explanation:
Given: [tex]T_{1}[/tex] = 237 K, [tex]P_{1}[/tex] = 101.33 kPa, [tex]V_{1}[/tex] = ?
[tex]T_{2}[/tex] = 300 K, [tex]P_{2}[/tex] = 151.99 kPa, [tex]V_{2}[/tex] = 0.50 L
Formula used is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L[/tex]
Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.
