Charles's and Gay-Lussac's Law practice

Question

1. A container holds 30.0 mL of nitrogen at 45° C. What will be its volume if the temperature increases to 65° C?

2. A sample of oxygen occupies a volume of 12 L at 1000° C. What will be the volume of oxygen when the temperature drops to 0.00° C?

3. A sample of hydrogen has an initial temperature of 60° C. When the temperature is lowered to 7.0° C, the volume of hydrogen becomes 24L . What was the initial volume of the hydrogen ?

4. 65L of chlorine at 35° C will occupy what volume at -75° C while the pressure remains constant?

5. I have 17 liters of gas in a piston at a temperature of 2800 C. If I cool the gas until the volume decreases to 8.5 liters, what will the temperature of the gas be?

6. If I have 65 liters of helium in a balloon at 150 C and increase the temperature of the balloon to 750 C, what will the new volume of the balloon be?

Gay Lussac’s Law 

1. Determine the pressure change when a constant volume of gas at 2.00 atm is heated from 18.0 °C to 40.0 °C.

2. A gas has a pressure of 0.390 atm at 50.0 °C. What is the pressure at standard temperature?

3. A gas has a pressure of 500.0 mmHg at 40.0 °C. What is the temperature at standard pressure?

4. If a gas is cooled from 500.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 550.0 mmHg?

5. A sample of nitrogen inside a metal container at 40.0 °C is placed inside an oven whose temperature is 60.0 °C. The pressure inside the container at 30.0 °C was at 5.00 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 °C?

6. If a gas in a closed container is pressurized from 12.0 atmospheres to 17.0 atmospheres and its original temperature was 35.0 °C, what would the final temperature of the gas be in degrees Celsius?

Answer 1

Charles's law: V₁/T₁ = V₂/T₂ at constant pressure. Every temperature must be in kelvin — add 273.15 to each Celsius value before doing anything else. That single step is where nearly all the errors in this set will come from.

1. 30.0 mL, 45 °C → 65 °C  (318.15 K → 338.15 K)
V₂ = 30.0 × (338.15/318.15) = 31.9 mL

2. 12 L, 1000 °C → 0.00 °C  (1273.15 K → 273.15 K)
V₂ = 12 × (273.15/1273.15) = 2.6 L

3. 60 °C → 7.0 °C, final volume 24 L, find the initial volume  (333.15 K → 280.15 K)
V₁ = 24 × (333.15/280.15) = 29 L

4. 65 L, 35 °C → −75 °C  (308.15 K → 198.15 K)
V₂ = 65 × (198.15/308.15) = 42 L

5. 17 L at 280 °C, cooled until the volume is 8.5 L — find the temperature.
Here you solve for T instead: T₂ = T₁ × (V₂/V₁) = 553.15 × (8.5/17) = 276.6 K = 3.4 °C
Halving the volume halves the absolute temperature — 553 K to 277 K — which is a drop from 280 °C to just above freezing, not to 140 °C. This problem exists to make that point.

6. 65 L, 15 °C → 75 °C  (288.15 K → 348.15 K)
V₂ = 65 × (348.15/288.15) = 79 L

A note on how questions 5 and 6 were typed

They appear as "2800 C", "150 C", and "750 C". That is a degree symbol that did not survive copying — the superscript became a zero. I have read them as 280 °C, 15 °C, and 75 °C, which is consistent with the other questions in the set. If your worksheet really does say 2800 °C, tell me and the arithmetic changes.

Check every answer for direction before moving on

Heating must increase volume, cooling must decrease it. Questions 1 and 6 got bigger; 2 and 4 got smaller; 3 works backwards from a cooled sample so the original volume is larger. All consistent. If any answer moves the wrong way, you have inverted the ratio — and that check costs nothing.

Your Gay-Lussac section was cut off after "a constant volume of gas at 2.00 at..." — post the rest and I will work those too. They use P₁/T₁ = P₂/T₂, the same structure with pressure in place of volume.