Problems #1 - 10

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**Notes:**

I used:

V_{1}/ T_{1}= V_{2}/ T_{2}

to set up the solution for the first few.

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

V_{1}T_{2}= V_{2}T_{1}

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:V

_{1}T_{2}= V_{2}T_{1}V

_{2}= (V_{1}T_{2}) / T_{1}<--- divided both sides by T_{1}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:V

_{1}T_{2}= V_{2}T_{1}V

_{2}= (V_{1}) [T_{2}/ T_{1}] <--- notice how I grouped the temperatures togetherx = (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:

V_{2}= (V_{1}) [T_{2}/ T_{1}]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:

V_{1}/ T_{1}= V_{2}/ T_{2}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 = V_{2}/ 294 KCalculate V

_{2}. The volume that "escapes" is V_{2}minus 2.05 L

Usually, a Charles' Law problem asks for what the volume is at the end (the V_{2} in this question) or at the start, before some temperature change. This question asks you for the __difference__ between V_{1} and V_{2}. It's not hard to solve, it's just that it doesn't get asked very often in a Charles' Law setting.

Go to Charles' Law Problems #11 - 25

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