ChemTeam: Charles' Law - Problems #1 (2024)

Charles' Law
Problems #1 - 10

Notes:

I used:

V₁ / T₁ = V₂ / T₂

to set up the solution for the first few.

Sometimes, you will see the symbolic equation in cross-multiplied form:

V₁T₂ = V₂T₁

I set up some solutions toward the end using various permutations of the cross-multiplied form.

In all the problems below, the pressure and the amount of gas are held constant.

Problem #1: Calculate the decrease in temperature (in Celsius) when 2.00 L at 21.0 °C is compressed to 1.00 L.

Solution:

(2.00 L) / 294.0 K) = (1.00 L) / (x)

cross multiply to get:

2x = 293

x = 147.0 K

Converting 147.0 K to Celsius, we find -126.0 °C, for a total decrease of 147.0 °C, from 21.0 °C to -126.0 °C.

Problem #2: 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?

Solution:

(600.0 mL) / (293.0) = (x) / (333.0 K)

x = 682 mL

See Also

Physlet Physics by Christian and Belloni: Problem 20.7Charles' Law CalculatorGas LawsWhat happens to the volume of the substance when the temperature increases?

Problem #3: A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?

Solution:

(900.0 mL) / (300.0 K) = (x) / (405.0 K)

x = 1215 mL

Problem #4: What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?

Solution:

(60.0 mL) / (306.0 K) = (x) / (278.00 K)

Cross multiply to get:

306x = 16680

x = 54.5 mL <--- that's the ending volume, which is NOT the answer

The volume decreases by 5.5 mL.

Problem #5: Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?

Solution:

In cross-multiplied form, it is this:

V₁T₂ = V₂T₁

V₂ = (V₁T₂) / T₁ <--- divided both sides by T₁

x = [(300.0 mL) (283.0 K)] / 290.0 K

Problem #6: A gas occupies 1.00 L at standard temperature. What is the volume at 333.0 °C?

Solution:

In cross-multiplied form, it is this:

V₁T₂ = V₂T₁

V₂ = (V₁) [T₂ / T₁] <--- notice how I grouped the temperatures together

x = (1.00 L) [(606.0 K) / (273.0 K)]

x = 2.22 L

Problem #7: At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?

Solution:

Two different set-ups:

(6.00 L) / (300.0 K) = (x) / (423.0 K)

or

(6.00 L) (423.0 K) = (x) (300.0 K)

Same answer:

x = 8.46 L

Problem #8: At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0 °C?

Solution:

From #6:

V₂ = (V₁) [T₂ / T₁]

x = (400.0 mL) [(400.0 K) / (498.0 K)

x = 321 mL

Here's the "traditional" way:

(400.0 mL) / (498.0 K) = (x) / (400.0 K)

Problem #9: At 210.0 °C a gas has a volume of 8.00 L. What is the volume of this gas at -23.0 °C?

Solution:

(8.00 L) / (483.0 K) = (x) / (250.0 K)

Note how you can have a negative Celsius temperature, but not a negative Kelvin temperature.

Problem #10: When the volume of a gas is changed from ___ mL to 852 mL, the temperature will change from 315 °C to 452 °C. What is the starting volume?

Solution:

Write Charles Law and substitute values in:

V₁ / T₁ = V₂ / T₂

x / 588 K = 852 mL / 725 K

(x) (725 K) = (852 mL) (588 K)

x = 691 mL

Note the large °C values, trying to get you to forget to add 273. Remember, only Kelvin temperatures are allowed in the calculations.

Bonus Problem: An open "empty" 2 L plastic pop container, which has an actual inside volume of 2.05 L, is removed from a refrigerator at 5 °C and allowed to warm up to 21 °C. What volume of air measured at 21 °C, will leave the container as it warms?

Solution:

2.05 L / 278 K = V₂ / 294 K

Calculate V₂. The volume that "escapes" is V₂ minus 2.05 L

Usually, a Charles' Law problem asks for what the volume is at the end (the V₂ in this question) or at the start, before some temperature change. This question asks you for the difference between V₁ and V₂. It's not hard to solve, it's just that it doesn't get asked very often in a Charles' Law setting.