11
vaibhav sharma
·2010-07-19 11:11:22
any 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
chessenthus
·2010-07-20 06:09:31
thank 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]?
1
gauraviscool
·2010-07-20 08:01:36
as 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]