Asked by Anonymous
A vessel containing 39.5 cm^3 of helium gas at 25°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 7.7cm^3 of ethanol, what was the final temperature of the helium?
Answers
Answered by
Jai
assuming ideal gas, we can use the formula for constant pressure:
V1/T1 = V2/T2
where
V = volume
T = temperature (in Kelvin)
note that to convert to Kelvin, K = C + 273
substituting,
39.5 / (25 + 273) = 7.7 / T2
now solve for T2. subtract 273 to convert it to deg C units
V1/T1 = V2/T2
where
V = volume
T = temperature (in Kelvin)
note that to convert to Kelvin, K = C + 273
substituting,
39.5 / (25 + 273) = 7.7 / T2
now solve for T2. subtract 273 to convert it to deg C units
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