A test tube contains 2.63x10-4 m3 of liquid carbon tetrachloride (β = 1240x10-6 (C°)-1) at a temperature of 80.0°C. The test tube and the carbon tetrachloride are cooled to a temperature of -10.0°C, which is above the freezing point of carbon tetrachloride. Find the volume of carbon tetrachloride in the test tube at -10.0°C.
A test tube contains 2.63x10-4 m3 of liquid carbon tetrachloride (β = 1240x10-6 (C°)-1) at a temperature of 80.0°C. The test tube and the carbon tetrachloride are cooled to a temperature of -10.0°C, which is above the freezing point of carbon tetrachloride. Find the volume of carbon tetrachloride in the test tube at -10.0°C.
The volume and temperature both are related to each other. This relationship is known as Charles law relationship. It given by following formula -
V1/T1 = V2/T2
Here,
V1 = initial volume of Carbon tetrachloride = 2.63×10-4 m3
V2 = final volume of Carbon tetrachloride = ?
T1 = initial temperature of Carbon tetrachloride = 80.0°C = 80.0+273 = 353 K
T2 = final temperature of Carbon tetrachloride = -10.0°C = -10.0+273 = 263 K
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