Charles' Law
Problems #11 - 25

I used V₁ / T₁ = V₂ / T₂ to set up the solution. I did not carry them through to a final answer

Problem #11: Carbon dioxide is usually formed when gasoline is burned. If 30.0 L of CO₂ is produced at a temperature of 1.00 x 10³ °C and allowed to reach room temperature (25.0 °C) without any pressure changes, what is the new volume of the carbon dioxide?

30.0 L / 1273 K = x / 298.0 K

Problem #12: A 600.0 mL sample of nitrogen is warmed from 77.0 °C to 86.0 °C. Find its new volume if the pressure remains constant.

600.0 mL / 350.0 K = x / 359.0 K

Problem #13: What volume change occurs to a 400.0 mL gas sample as the temperature increases from 22.0 °C to 30.0 °C?

400.0 mL / 295.0 K = x / 303.0 K

Problem #14: A gas syringe contains 56.05 milliliters of a gas at 315.1 K. Determine the volume that the gas will occupy if the temperature is increased to 380.5 K

56.05 mL / 315.1 K = x / 380.5 K

Problem #15: A gas syringe contains 42.3 milliliters of a gas at 98.15 °C. Determine the volume that the gas will occupy if the temperature is decreased to -18.50 °C.

42.3 mL / 371.15 K = x / 254.50 K

Problem #16: When the temperature of a gas decreases (with pressure and amount held constant), does the volume increase or decrease?

Decrease.

Problem #17: If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by ____.

A factor of 2.

Problem #18: Solve the Charles' Law equation for V₂.

V₂ = (V₁ times T₂) / T₁

or

V₂ = V₁ times (T₂ / T₁)

Problem #19: Charles' Law deals with what quantities?

a. pressure/temperature
b. pressure/volume
c. volume/temperature
d. volume/temperature/pressure

Problem #20: If 540.0 mL of nitrogen at 0.00 °C is heated to a temperature of 100.0 °C what will be the new volume of the gas?

540.0 mL / 273.0 K = x / 373.0 K

Problem #21: A balloon has a volume of 2500.0 mL on a day when the temperature is 30.0 °C. If the temperature at night falls to 10.0 °C, what will be the volume of the balloon if the pressure remains constant?

2500.0 mL / 303.0 = x / 283.0 K

Problem #22: When 50.0 liters of oxygen at 20.0 °C is compressed to 5.00 liters, what must the new temperature be to maintain constant pressure?

50.0 L / 293.0 K = 5.00 L / x

Problem #23: If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the new temperature be to maintain constant pressure?

15.0 L / 298.0 K = 45.0 L / x

Problem #24:3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?

3.50 liters / 1000.0 K = x / 426.0 K

Problem #25: If your volume-temperature constant is 0.432 mL/K, what would the volume of your sample be at 35.0 °C

In my Charles' Law discussion, I gave Charles' law as this:

V / T = k

Here's the set up to solve the problem:

x / 308 K = 0.432 mL/K

x = 133 mL

It is unusual to see a question that uses V / T = k. The much, much more common equation for Charles' Law problem solving is V₁ / T₁ = V₂ / T₂.

Bonus Problem: In an air-conditioned room at 19.0 #&176;C, a spherical balloon had the diameter of 50.0 cm. When taken outside on a hot summer day, the balloon expanded to 51.0 cm in diameter. What was the temperature outside in degrees Celsius? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.

Solution #1:

1) Since the pressure and amount of gas are constant, this problem becomes a Charles Law problem:

V₁ / T₁ = V₂ / T₂

solving for T₂, we have:

T₂ = V₂T₁ / V₁

2) Given the formula for volume of a sphere = (⁴⁄₃) π r³, we substitue and solve for T₂:

T₂ = [(⁴⁄₃) π r₂³) (T₁)] / [(⁴⁄₃) π r₁³]

T₂ = [(r₂³) (T₁)] / r₁³

T₂ = [(25.5 cm)³ (292.15 K)] / (25.0 cm)³

T₂ = 310.03 K

Converted to Celsius and three sig figs, 36.9 °C

Solution #2:

Volume of a sphere is proportional to the diameter cubed

The increase in volume to 51 cm from 50 cm = (51 / 50)³ = 1.0612 of the original volume

V₁ / T₁ = V₂ / T₂

V₁ / 292 K = (1.0612 * V₁) / T₂

T₂ = 309.9 K = 36.9 °C