A sample of nitrogen gas kept in a container of volume 2.3L and at a temp of 305K exerts a pressure of 4.7 atm. Calculate the number of moles of gas present.
2007-03-29 11:55:14 · 5 answers · asked by Me 3 in Science & Mathematics ➔ Chemistry
Use the Ideal Gas Law
PV = nRT
P = Pressure, in atmospheres
V = Volume, in Liters
n = number of moles of gas, in moles
R = 0.082 [L*atm/mol*K], the gas constant, a given
T = temperature, in Kelvins
P = 4.7 atm
V = 2.3 L
T = 305K
PV = nRT ---> n = (PV/RT)
n = ((4.7 [atm])*(2.3 [L])) / (0.082 [L*atm/mol*K] * 305[K])
= 0.432 moles
2007-03-29 12:42:15 · answer #1 · answered by Anonymous · 0⤊ 0⤋
PV = n RT
P = Pressure (atm)
V = Volume (L)
n = moles
R = gas constant = 0.082057 L atm mol-1 K-1
T = temperature (kelvins)
Look at the article, memorize pv = nrt. Remember to use the proper units and the proper gas constant that goes with the units!
2.3 X 4.7 = n (.082057)( 305)
solve for n, i hate the computer calculator
2007-03-29 12:01:37 · answer #2 · answered by spidermilk666 6 · 0⤊ 0⤋
Yes use PV=nRT which is the Ideal Gas Law. I think you have to make sure the units used are the same as used for R the Gas Constant.
2007-03-29 12:05:30 · answer #3 · answered by pschroeter 5 · 0⤊ 0⤋
keep in mind the Avogadro No. The Avogadro huge type is the shape of molecules in a unmarried mole of a substance. The Avogadro No. is 6.022 x 10^23 in a unmarried mole. In ammonia. The Mr(NH3) = 14 + (3 x a million) = 17 mol(NH3) = 3.7 / 17 = 0.2176 So quite of having one mole we purely have approx. 0.2176 moles. So utilising the Avogadro No.and multiplying by technique of 0.2176 for this reason 6.022 x 10^23 x 0.2176 = a million.31 x 10^23 (as required).
2016-12-03 00:07:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
PV=nRT
4.7 * 2.3 = n * 0.0821 * 305
Solve for n (and check R -- my memory is rusty)
2007-03-29 11:58:03 · answer #5 · answered by tedfischer17 3 · 0⤊ 0⤋