3.01 atm
According to Gay-Lussac's law, the pressure (P) of a given mass of gas when volume is left constant is directly proportional to the kelvin temperature (T) of the gas. i.e
P ∝ T
P = kT
Where;
k = proportionality constant
=> [tex]\frac{P}{T}[/tex] = k
=> [tex]\frac{P_{1} }{T_{1} }[/tex] = [tex]\frac{P_{2} }{T_{2} }[/tex] -------------------(i)
Where;
[tex]P_{1}[/tex] and [tex]P_{2}[/tex] are the initial and final pressures of the gas
[tex]T_{1}[/tex] and [tex]T_{2}[/tex] are the initial and final temperatures of the gas
From the question;
[tex]P_{1}[/tex] = 2.48 atm
[tex]P_{2}[/tex] = ?
[tex]T_{1}[/tex] = 21°C = (21 + 273)K = 294K
[tex]T_{2}[/tex] = 84°C = (84 + 273)K = 357K
Substitute these values into equation (i) as follows;
=> [tex]\frac{2.48}{294}[/tex] = [tex]\frac{P_{2} }{357}[/tex]
Solve for [tex]P_{2}[/tex]
[tex]P_{2}[/tex] = [tex]\frac{2.48 X 357}{294}[/tex]
[tex]P_{2}[/tex] = 3.01 atm
Therefore the pressure will be 3.01 atm
