The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
V = \(\frac{k}{M}\)
V = \(\frac{k}{M^{\frac{1}{2}}}\)
V = \(kM^2\)
V = \((\frac{k}{M})^{\frac{1}{2}}\)
Explanation
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My school the above provided answer is wrong
V=√(3RT/M)----------(i)
M in the above formula does not stand for the mass of the gas it stands for the molar mass of the gas
we already know that the molar mass (M)of a gas =mass(m)/moles(n)--------(ii)
hence ,substituting (ii) in to eqn (i) we have
V=√(3RT/m/n)= √(3nRT/m)
from the above expression it is clear that
V=,(k/m)^1/2
When we talk about the velocity of a gas , you can liken it to rate of diffusion of the gas, R. Don't forget that rate of diffusion can be observed in 2 ways: 1.the distance the gas travels per time and 2.the volume of gas dissipated per time.
The first method can be likend to velocity since it is distance covered per time {don't forget that anything 'rate' means per time}. Using Grahams law of diffusion which states that the rate of diffusion of a gas is inversely proportional to the square root of it's molar mass; We'll have:
R = K/√(M.M), since we likened rate of diffusion to be velocity (since they have the same unit for case 1), we'll replace the R there with V for velocity; V = K/√(M.M). Now we know that molar mass (M.M) is equal to mass/no. of moles. Substituting that into V = K/√(M.M), we'll have:
V = K/√(M/n), where M is mass & n is no. of moles. Then V = K√(n/M), if the no. of moles is constant, then V = K/√M which is the same as
V = K/(M)½. I hope this helps.
Graham law of diffusion
The rate or diffusion of a gas is inversely proportional to the squareroot of its density or RMM
We can assume that the rate here is the velocity of the gas
therefore, V = 1/sqrt(M)
Kinetic energy = 1/2 * mass * volume * volume
KE = 1/2m(v)^2 ..........(1)
from(1)
v = (2KE/m)^(1/2)
Since two and KE are constant values
we can say that v = k/(m)^1/2
