Charles' Law
Problems #1 - 10

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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


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.


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