Gay-Lussac Problems
P./T1 = P2/T2 - Remember, T is in KELVIN, add 273 to any temp given to you in °C to convert
1. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0
°C to 30.0 °C.
2. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its
original temperature was 25.0 °C, what would the final temperature of the gas be?
3. A gas has a pressure of 0.35 atm at 50.0 °C. What is the pressure at standard temperature?
Boyle's law Drobleme

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:

=

Putting the values in equation:

P2 =

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,

=

putting values in the equation:

T2 =

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:

=

P2 =

P2    = 0.29 atm

0.29 atm is the pressure at standard temperature.

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