Answer: The new pressure is 280 kPa.
Explanation:
Given: [tex]V_{1}[/tex] = 400.0 mL, [tex]T_{1} = 20^{o}C[/tex] , [tex]P_{1}[/tex] = 70.0 kPa
[tex]V_{2}[/tex] = 200.0 mL, [tex]T_{2} = 40.0^{o}C[/tex], [tex]P_{2}[/tex] = ?
Now, combined gas law is used to calculate the new pressure 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{70.0 kPa \times 400.0 mL}{20^{o}C} = \frac{P_{2} \times 200.0 mL}{40.0^{o}C}\\P_{2} = 280 kPa[/tex]
Thus, we can conclude that the new pressure is 280 kPa.
