8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

Question

Cameron Andrews · Accepted Answer

The gas' new volume will be #V_2 = 7.28# L.

Charles' law is applied simply to this problem.

#V_1/T_1 = V_2/T_2#,

which states that the volume of a gas is directly proportional to its temperature(measured in #K# - very important).

So, we have an initial volume equal to #V_1 = 8.00L#, an initial temperature of #T_1 = (273.15+60) = 333.15K#, and a final temperature of #T_2 = (273.15 + 30) = 303.15K#, which gets us

#V_2 = T_2/T_1 * V_1 = 303.15/333.15 * 8.00 = 7.28L# -> a decrease in temperature is correlated with a decrease in volume, just as expected.

But take note of what would have occurred if degrees Celsius had been used in place of K:

#V_2 = 30/60 * 8.00 = 4.00 L#, a result significantly different...

Avery Adams · Answer

undefined To find the volume of the gas at 30.0°C, you can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature at constant pressure.

V1 / T1 = V2 / T2

Given:
V1 = 8.00 L
T1 = 60.0°C = 333.15 K (converted to Kelvin)
T2 = 30.0°C = 303.15 K (converted to Kelvin)

Solve for V2:

V2 = (V1 * T2) / T1

V2 = (8.00 L * 303.15 K) / 333.15 K

V2 ≈ 7.29 L

So, the volume of the gas upon cooling to 30.0°C will be approximately 7.29 liters.