A vessel containing 39.5 cm3 of helium gas at 25 degrees C and 106kPa was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required 7.7cm3 of ethanol, what was the final temperature of the helium?

Hi Corey,

The best thing to do in this type of a problem is write out your known and unknown variables first. You have to make sure that your temperature is in degrees Kelvin, and that your volume is in Liters. Pressure usually must be in atmospheres, but we don't need to change it since it remains constant.

V

_{1}=39.5cm^{3}= 0.0395LT

_{1}=25ÂșC = 298KP

_{1 }= 106kPaV

_{2 }=39.5cm^{3}-7.7cm^{3}= 31.8cm^{3 }=0.0318L (You subtract the 7.7cm^{3}volume of ethanol because this is how much the original volume shrunk by)T

_{2}=?P2=106kPa (because they said you keep pressure constant in the problem)

Since the pressure stays the same, it means you need to use Charles's Law to solve for T

_{2.}V

_{1}/T_{1}=V_{2}/T_{2}is the equation for Charles's Law. (rearrange to get T_{2}= V_{2}T_{1/}V_{1})Plug in the values you have up above and you will get: T

_{2}=0.0318LX298K/0.0395L = 239.9KI hope that helps,

Cody