Answer:
The answer is e.
Explanation:
We take:
[tex]T_{1}=253K[/tex]
[tex]V_{1}=2.00 l[/tex]
[tex]P_{1}=0.14atm[/tex]
[tex]V_{2}=4.50 l[/tex]
[tex]T_{2}=330K[/tex]
Taking the gas as an ideal gas, we can use the ideal gas law:
[tex]\frac{PV}{nRT}[/tex]
⇒ [tex]n=\frac{P_{1}V_{1}}{RT_{1}}[/tex] ⇒ [tex]n=0.013mol[/tex]
Then:
[tex]P_{2}=\frac{nRT_{2}}{V_{2}}[/tex] ⇒ [tex]P_{2}=0.0811 atm[/tex]
Taking R=0.08205atmL/molK.
