An air thermometer can be constructed by using a glass bubble attached to a piece of small-diamter glass tubing. The tubing contains a small amount of colored water that rises when the temperature increases and the trapped air expands. You want a 0.20 cm3 change in volume to equal a 1° C change in temperature. What total volume of air at 20° C should be trapped in the apparatus below the liquid?

In a certain experiment a sample of Helium in a vacuum system was compressed at 25 degrees C from a volume of 200.0 mL to a volume of 0.240 mL where its pressure was found to be 30.0 mm Hg. What was the orginal pressure of the Helium?

To determine the total volume of air at 20°C that should be trapped in the apparatus below the liquid, we need to understand the relationship between volume and temperature. According to the provided information, we want a 0.20 cm³ change in volume to correspond to a 1°C change in temperature.

The proportional relationship between volume and temperature is described by Charles's Law, which states that the volume of a gas is directly proportional to its temperature, assuming constant pressure:

V₁/T₁ = V₂/T₂

Where:
V₁ = initial volume
T₁ = initial temperature
V₂ = final volume
T₂ = final temperature

In this case, we know that the change in volume (ΔV) is 0.20 cm³, and the change in temperature (ΔT) is 1°C. We can set up the equation using the given values:

(ΔV) / (ΔT) = (V₂ - V₁) / (T₂ - T₁)

Since the effect is due to only a change in temperature, we can assume the initial volume (V₁) is equal to the final volume (V₂). Thus, the equation becomes:

(ΔV) / (ΔT) = (V₂ - V₁) / (T₂ - T₁) = (V₂ - V₁) / 1

Simplifying further:

ΔV = V₂ - V₁

Now, we substitute the given values: ΔV = 0.20 cm³ and ΔT = 1°C.

0.20 cm³ = V₂ - V₁

We still need to find the volume of air at 20°C, so we need to determine the initial volume (V₁) and the final volume (V₂).

Since the question asks for the total volume of air at 20°C that should be trapped in the apparatus below the liquid, we can assume that the final temperature (T₂) is 20°C. Therefore, T₂ = 20°C.

Now, we can rearrange the equation and solve for V₁:

V₁ = V₂ - ΔV

Substituting the known values:

V₁ = 20 cm³ - 0.20 cm³
V₁ = 19.80 cm³

Therefore, the total volume of air at 20°C that should be trapped in the apparatus below the liquid is 19.80 cm³.