Explanation:
The given data is as follows.
    [tex]V_{1}[/tex] = 4.60 L,     [tex]P_{1}[/tex] = 365 mm Hg
    [tex]V_{2}[/tex] = 2.60 L,    [tex]P_{2}[/tex] = ?
    [tex]T_{1}[/tex] = (20 + 273) K = 293 K,    [tex]T_{2}[/tex] = (36 + 273) K = 309 K
Since, number of moles of gas are equal so, according to ideal gas equation:
     [tex]\frac{P_{1}V_{1}}{T_{1}}[/tex] = [tex]\frac{P_{2}V_{2}}{T_{2}}[/tex]
     [tex]\frac{365 mm Hg \times 4.60 L}{293 K}[/tex] = [tex]\frac{P_{2} \times 2.60 L}{309 K}[/tex]
     [tex]P_{2}[/tex] = 681.03 mm Hg
Thus, we can conclude that new pressure [tex]P_{2}[/tex] is 681.03 mm Hg.
