sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

- 81.5°C

Explanation:

Data obtained from the question include:

V1 (initial volume) = 3.20 L

T1 (initial temperature) = 125 °C = 125 + 273 = 398K

V2 (final volume) = 1.54 L

T2 (final temperature) =.?

At constant pressure indicates that the gas is obeying Charles' law. Using the Charles' law equation V1/T1 = V2/T2, the final temperature of the gas can be obtained as illustrated below:

V1/T1 = V2/T2

3.2/398 = 1.54/T2

Cross multiply to express in linear form

3.2 x T2 = 398 x 1.54

Divide both side by 3.2

T2 = (398 x 1.54) / 3.2

T2 = 191.5K

Now let us convert 191.5K to celsius temperature. This is illustrated below

°C = K - 273

°C = 191.5 - 273

°C = - 81.5°C

Therefore the gas will occupy 1.54L at a temperature of - 81.5°C

-81.5 degrees C or 191.5 K

Explanation:

We want to use Charles' gas law: V/T = V/T

Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.

Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:

3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K

If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = - 81.5 degrees C