The pressure inside a hydrogen-filled container was 2.10 atm at 21 ?C . What wou
ID: 839240 • Letter: T
Question
The pressure inside a hydrogen-filled container was 2.10 atm at 21 ?C . What would the pressure be if the container was heated to 85?C ?
Express your answer with the appropriate units.
At standard temperature and pressure (0 ?C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L . What volume would the same amount of gas occupy at the same pressure and 40?C ?
Express your answer with the appropriate units.
In an air-conditioned room at 19.0 ?C a spherical balloon had the diameter of 50.0 cm . When taken outside on a hot summer day, the balloon expanded to 51.0cm in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
Express your answer with the appropriate units.
A cylinder with a movable piston contains 2.00 g of helium, He , at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.30L ? (The temperature was held constant.)
Express your answer with the appropriate units.
A certain gas is present in a 10.0L cylinder at 4.0atm pressure. If the pressure is increased to 8.0atm the volume of the gas decreases to 5.0L . Find the two constants ki , the initial value of k , and kf , the final value of k , to verify whether the gas obeys Boyle
Explanation / Answer
PV = nRT... R is a contant...
so PV/nT = constant...
P1V1 / n1T1 = P2V2 / n2T2...P is constant.T is constant.. so
V1 / n1 = V2 / n2
since n = mass / mw
V1 / (mass 1 / mw) = V2 / (mass 2 / mw)
and since molecular mass = constant
V1 / mass 1 = V2 / mass 2
mass 2 = V2 x mass 1 / V1 = 4.30L x 2.00 g / 2.00 L = 4.30 g
mass 2 - mass 1 = 4.30 g - 2.00 g = 2.30 g
ie.. 2.30 g of He was added.
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