Respuesta :
According to Gay-Lussac's law of pressure–temperature: pressure (P) of gas is directly proportional to temperature (T).
The formula is given by:
[tex]\frac{P}{T}=k (constant))[/tex] or,
[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex] (1)
[tex]P_{1}[/tex] (Pressure of gas) =1.26 bar
[tex]T_{1}[/tex] (temperature of gas) =[tex]58.0^{o}C[/tex]
[tex]P_{2}[/tex] (Pressure of gas) =?
[tex]T_{2}[/tex] (temperature of gas) =[tex]118.0^{o}C[/tex]
Substitute the above values in formula (1)
[tex]\frac{1.26 bar}{58.0^{o}C}=\frac{P_{2}}{118.0^{o}C}[/tex]
[tex]\frac{1.26 bar}{58.0^{o}C}=\frac{P_{2}}{118.0^{o}C}[/tex]
Convert [tex]^{o}C[/tex] into kelvin
[tex]\frac{1.26 bar}{(58.0^{o}C +273 )}\times (118.0^{o}C+273 )= P_{2}[/tex]
[tex]P_{2} =\frac{1.26 bar}{331 K}\times (391 K) [/tex]
= [tex]1.488 bar \simeq 1.49 bar[/tex]
Thus, pressure of the gas at [tex]118.0^{o}C[/tex] is 1.49 bar.
