Answer:
The change in volume is -5.5 mL (a decrease in volume of 5.5 mL) when 60.0 mL of gas is cooled from 33.0°C to 5.00°C.
Explanation:
To calculate the change in volume, we need to use the ideal gas law equation:
V1/T1 = V2/T2
where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
V1 = 60.0 mL
T1 = 33.0°C = 33.0 + 273.15 = 306.15 K
T2 = 5.00°C = 5.00 + 273.15 = 278.15 K
Now we can calculate V2, the final volume:
V1/T1 = V2/T2
(60.0 mL) / (306.15 K) = V2 / (278.15 K)
Cross-multiplying and solving for V2:
V2 = (60.0 mL) * (278.15 K) / (306.15 K)
V2 = 54.5 mL
The final volume, V2, is 54.5 mL.
To find the change in volume, we subtract the initial volume from the final volume:
Change in volume = V2 - V1
Change in volume = 54.5 mL - 60.0 mL
Change in volume = -5.5 mL
Therefore, the change in volume is -5.5 mL (a decrease in volume of 5.5 mL) when 60.0 mL of gas is cooled from 33.0°C to 5.00°C.