Question 1 (5 points) Charles\'s law states that: The total pressure of a mixtur

Question

Question 1 (5 points)

Charles's law states that:

The total pressure of a mixture of gases is the simple sum of the partial pressure of all of the gaseous compounds.

The volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature.

The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure.

Equal amounts of gases occupy the same volume at constant temperature and pressure.

The rates of effusion of gases are inversely proportional to the square roots of their molar masses.

Question 2 (5 points)

A gas sample is held at constant pressure. The gas occupies 2.78 L of volume when the temperature is 21.2°C. Determine the temperature (in °C) at which the volume of the gas is 2.16 L.

Your Answer:Question 2 options:

Question 3 (5 points)

The volume of a baloon at 25.5 °C is 2.23 L, what is the volume at 37.9 °C assuming pressure is constant.

Your Answer:Question 3 options:

The total pressure of a mixture of gases is the simple sum of the partial pressure of all of the gaseous compounds.

The volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature.

The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure.

Equal amounts of gases occupy the same volume at constant temperature and pressure.

The rates of effusion of gases are inversely proportional to the square roots of their molar masses.

Explanation / Answer

Q 1) Charles's law states that at constant pressure the volume of gas is directly proportional the temperature.

So answer is - The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure.

Q 2) Given, V1 = 2.78 L , T1 =21.2oC + 273 = 294.2 K

V2 = 2.16 L, T2 = ?

We know Charles's law

V1/T1 = V2/T2

So, T2 = V2*T1 /V1

            = 2.16 L * 294.2 K / 2.78 L

             = 228.6 K

so temperature in the degree Celsius = 228.6 – 273 = -44.4oC

Q 3) Given, V1 = 2.23 L , T1 =25.5oC + 273 = 298.5 K

V2 = ? L, T2 = 37.9 oC + 273 = 310.9 K

We know Charles's law

V1/T1 = V2/T2

So, V2 = V1*T2 /T1

            = 2.23 L * 310.9 K / 298.5 K

             = 2.32 L

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