1) 1.03 atm is the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.
2) 318.02 K will be the final temperature of the gas.
3) 0.29 atm is the pressure at standard temperature of 273.15 K.
Explanation:
a) Given that formula used is Gay-Lussac Law:
volume = constant
pressure P1 = 1 atm
temperature T1 = 20 Degrees or 293.15 K
T2 = 30 Degrees T2 = 303.15
P2 = ?
since volume remains constant, equation used:
[tex]\frac{P1}{T1}[/tex]= [tex]\frac{P2}{T2}[/tex]
Putting the values in equation:
P2 = [tex]\frac{1 X 303.15}{293.15}[/tex]
P2 = 1.03 atm. is the atmospheric change.
2) Given:
P1 = 15 atm
P2 = 16 atm
T1 = 25 Degrees or 298.15 K
T2 = ?
using the formula,
[tex]\frac{P1}{T1}[/tex]= [tex]\frac{P2}{T2}[/tex]
putting values in the equation:
T2 = [tex]\frac{298.15 X 16}{15}[/tex]
T2 = 318.02 K is the final temperature of the gas.
3) given:
P1 = 0.35 atm.
T1 = 50 Degrees or 323.15 K
P2 = ?
T2 = 273.15 K
putting the values in the equation:
[tex]\frac{P1}{T1}[/tex]= [tex]\frac{P2}{T2}[/tex]
P2 = [tex]\frac{0.35 X 273.15}{323.15}[/tex]
P2 = 0.29 atm
0.29 atm is the pressure at standard temperature.
