If a sample of gas occupies a volume of 2.50 liters at 25.0 degrees C, would the volume be larger or smaller if the temperature drops to -5.0 degrees C?
Same pressure? Charles'law...
V1/T1=V2/T2
temps in Kelvins
So would it be:
2.50 L/298 K = V/268 k
To determine if the volume would be larger or smaller when the temperature drops from 25.0 degrees Celsius to -5.0 degrees Celsius, we need to apply the gas laws, specifically Charles's Law.
Charles's Law states that, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin.
To compare the volumes at two different temperatures, we need to convert both temperatures to Kelvin. The Kelvin temperature scale does not use the degree symbol, so we add 273.15 to convert from Celsius to Kelvin.
Given:
Initial volume (V1) = 2.50 liters
Initial temperature (T1) = 25.0 degrees Celsius = 25.0 + 273.15 = 298.15 K
Final temperature (T2) = -5.0 degrees Celsius = -5.0 + 273.15 = 268.15 K
Next, we can set up the ratio of the initial volume to the final volume using Charles's Law:
V1/T1 = V2/T2
Using the known values, we can write:
2.50 L / 298.15 K = V2 / 268.15 K
Now, we can solve for V2:
V2 = (2.50 L / 298.15 K) * 268.15 K
≈ 2.25 L
Therefore, the volume would be smaller (approximately 2.25 liters) when the temperature drops to -5.0 degrees Celsius.