A vessel containing 39.5 cm3 of helium gas at 25°C and 106 kPa 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 17.8 cm3 of ethanol, what was the final temperature of the helium?
Didn't I do this earlier when you piggy backed it onto another post?
To determine the final temperature of the helium gas, we can use the ideal gas law equation:
PV = nRT
where:
P is the pressure of the gas (in this case, 106 kPa)
V is the volume of the gas (initial volume of helium = 39.5 cm³)
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature of the gas (we need to find this)
Since the pressure is kept constant during this process, we can rewrite the equation as:
V1/T1 = V2/T2
where:
V1 is the initial volume of helium (39.5 cm³)
T1 is the initial temperature of helium (given as 25°C or 298 K)
V2 is the final volume of the gas (39.5 cm³ + 17.8 cm³)
T2 is the final temperature of the helium (we need to find this)
Rearranging the equation to solve for T2:
T2 = (V2 * T1) / V1
Plugging in the given values:
T2 = ((39.5 cm³ + 17.8 cm³) * 298 K) / 39.5 cm³
T2 = (57.3 cm³ * 298 K) / 39.5 cm³
Calculating the result:
T2 = 4322 K / 39.5 cm³
T2 = 109.62 K
Therefore, the final temperature of the helium gas is approximately 109.62 K.