1thank you for the answer.But I could not understand one thing.... can you tell me how the dimensions of α become [MLT-2]?In other words what are the dimensions of Boltzmann constant and temperature so that the final answer becomes [MLT-2]?
In the relation;
P=αβe-αZkθ,where P is pressure,Z is the distance,k is Boltzmann's constant and θ is the temperature.The dimensional formula for β will be:
a)[L2]
b)[ML2T]
c)[MT-1];here T is raised to the power -1
d)[L2T-1];here T is raised to the power -1
What is the dimensional formula for Boltzmann's constant?
Please expain step-wise...
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3 Answers
11any term which is in power of e is dimensionless so
aZk∂(i have used ∂ in place of theta)=dimensionless
so dimension of a = dimension of k∂Z=[MLT -2]
dimension of ab=dimension of pressure
dimension of b = dimension of adimension of pressure
=[L2]
1
1as arguments of exponential functions is dimensionless therefore
[aZ/k∂] = [M0L0T0]
so [a]=[k∂/Z]
now [P]=[a/b]
so =[a/P]=[k∂/ZP]
now, dimensions of k∂ r that of energy
hence =[ML2T-2/LML-1T-2] = [L2]