Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 2.36 L, [tex]T_{1}[/tex] = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K,
[tex]P_{1}[/tex] = 10.0 atm, [tex]V_{2}[/tex] = 7.79 L,
[tex]T_{2}[/tex] = ?, [tex]P_{2}[/tex] = 5.56 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Hence, putting the given values into the above formula to calculate the value of final temperature as follows.
[tex]\frac{10.0 atm \times 2.36 L}{298 K} = \frac{5.56 atm \times 7.79 L}{T_{2}}[/tex]
[tex]T_{2}[/tex] = [tex]\frac{43.3124}{0.0792}[/tex] K
= 546.87 K
Thus, we can conclude that the final temperature is 546.87 K.