Gay-Lussac's Law
Problems #1 - 10

Five Examples

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Problem #1: A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 °C?

Solution:

P1   P2
––– = –––
T1   T2

3.00   x
––– = –––
293   323

Solution technique: cross-multiply and divide.

x = 3.31 atm (to three sig figs)

Note: you will see set ups (especially in gas laws) that simply omit all the units in the solution. If you do that on a homework problem or test, you may get a deduction. It's not laziness on the part of the person writing the solution, it's simply assuming the reader knows what the units are and how they cancel out to leave the final unit.

Many times, you (as the student) are not allowed that luxury.


Problem #2: Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.

Solution:

P1 / T1 = P2 / T2

1.00 atm / 20.0 = x / 30.0

x = 1.50 atm

Seems pretty easy. But, it's wrong! Why? I used Celsius rather than Kelvin. Here's the correct solution:

1.00 atm / 293 = x / 303

x = 1.03 atm

Makes bit of a difference, doesn't it?


Problem #3: A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature?

Solution:

0.370 atm   x
–––––––– = –––––
323 K   273 K

x = 0.313 atm (to three sig figs)

I put the units into this problem solution where, up above, I did not. Your teacher may deduct if you do not include the units in a problem solution on a homework problem or on a test.


Problem #4: A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure?

Solution:

699.0 mmHg   760.0 mmHg
–––––––– = ––––––––––
313 K   x

x = 340. K (or 67.0 °C. to three sig figs)

Note that the problem did not specify what temperature unit to use. Usually, the expectation is that the answer be given in the same unit as used in the problem (in this case, degrees Celsius). If you notice this done on a test question, consider asking the teacher what he/she wants done (or just supply both values).


Problem #5: If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mmHg?

Solution:

P1   P2
––– = –––
T1   T2

750.0 mmHg   x
––––––––––– = ––––––––
323.0 K   273.15 K

(750.0 mmHg) (273.15 K) = (323.0 K) (x)

x = 634.2 mmHg (to four sig figs)


Problem #6: 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 in degrees Celsius?

Solution:

15.0 atm   16.0 atm
–––––––– = ––––––––––
298 K   x

x = 44.9 °C. to three sig figs)

Note: convert from K to °C and then round off.


Problem #7: A 30.0 L sample of nitrogen inside a metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?

Solution:

P1 / T1 = P2 / T2

3.00 atm / 293.0 K = x / 323.0 K

x = 3.31 atm (to three sig figs)

Notice the inclusion of a volume in the problem. It was put there as a deliberate distraction. You get all concerned about what happens to the volume and you miss that it's a metal container, which we can assume has a fixed volume.


Problem #8: A sample of gas at 3.00 x 103 mmHg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank?

Solution:

P1 / T1 = P2 / T2

P1T2 = P2T1

P2 = (P1T2) / T1

P2 = [(3000 mmHg) (273 K)] / 773 K

The answer should be determined to three significant figures.


Problem #9: The temperature of a sample of gas in a steel container at 30.0 kPa is increased from -100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank.

Solution:

P2 = (P1T2) / T1

P2 = [(30.0 kPa) (1273 K)] / 173 K

The answer should be determined to three significant figures.


Problem #10: Calculate the final pressure inside a scuba tank after it cools from 1.00 x 103 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm.

Solution:

P2 = [(130.0 atm) (298 K)] / 1273 K

The answer should be determined to three significant figures.


Five Examples

KMT & Gas Laws Menu