Answer: The moles of gas present in the cylinder is 0.34 moles.
Explanation:
Given: [tex]P_{1}[/tex] = 2.7 atm, [tex]V_{1}[/tex] = 3.1 L, [tex]T_{1}[/tex] = 300 K
[tex]P_{2}[/tex] = ?, [tex]V_{2}[/tex] = 9.4 L, [tex]T_{2}[/tex] = 610 K
Formula used to calculate the final temperature 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{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm[/tex]
Now, moles present upon heating the cylinder are as follows.
[tex]P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol[/tex]
Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.
