Asked by Robotic
Calculate the mass of 3.7dm^3 of hydrogen at 21.5°c and a pressure of 102Pa?
Answers
Answered by
R_scott
p * v = n * r * t
you want n (number of moles)
t should be in ºK
make sure the value of r matches the other units
you want n (number of moles)
t should be in ºK
make sure the value of r matches the other units
Answered by
oobleck
PV = nRT
That is, PV/nT = R, a constant.
Now, you know that 1 mole occupies 22.4L = 22.4 dm^3 at STP (1atm = 101.325kPa, 273.15ºK)
So, you want to find n (moles) such that
(102)(3.7)/(n*(273.15+21.5)) = (101.325)(22.4)/(1*273.15)
n = 0.154 moles ≈ 0.30 g of H2
sanity check: the pressure and temperature are not too far off from STP.
the volume is about 1/7 of 22.4L, so we expect about 1/7 mole of H2.
Looks good.
That is, PV/nT = R, a constant.
Now, you know that 1 mole occupies 22.4L = 22.4 dm^3 at STP (1atm = 101.325kPa, 273.15ºK)
So, you want to find n (moles) such that
(102)(3.7)/(n*(273.15+21.5)) = (101.325)(22.4)/(1*273.15)
n = 0.154 moles ≈ 0.30 g of H2
sanity check: the pressure and temperature are not too far off from STP.
the volume is about 1/7 of 22.4L, so we expect about 1/7 mole of H2.
Looks good.
Answered by
R_scott
the pressure units in the problem are Pa ... not kPa
will make a difference
will make a difference
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