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sneha sunil raikar Grade: 12
        the van der waals constants for a gas are:4 litre square atm molsquare inverse., b=.04 lit mol inverse .its boyle temperature is roughly-
8 years ago

Answers : (2)

AskiitianExpert Shine
10 Points
										

Hi


In thermodynamics, the Boyle temperature is defined as the temperature for which the second virial coefficient, B2(T) vanishes, i.e. B2(T) = 0. Since higher order virial coefficients are generally much smaller than the second coefficient, the gas tends to behave as an ideal gas over a wider range of pressures when the temperature reaches the Boyle temperature. In any case, when the pressures are low, the second virial coefficient will be the only relevant one because the remaining concern terms of higher order on the pressure. We then have dZ / dp = 0 at p = 0, where Z is the compression factor.


So, if the options are available then one can determine that at which temp, the B(t) tends to zero.

8 years ago
AskiitianExpert Shine
10 Points
										

Hi


At low pressures, all gases behave ideally, that is, they follow the Ideal Gas Law



pVm = RT



where p = pressure, Vm = molar volume, T = temperature and R is the Gas Constant.


However, at higher pressures gases deviate from the Ideal Gas Law, following a linear expansion of the form



pVm = RT{1 + B´p + C´p2 + ...}



sometimes stated as



pVm = RT{1 + B/Vm + C/Vm2 + ...}



The values of the virial coefficients B, C, ... depend on temperature, and many gases have a particular temperature at which B = B´ = 0. This temperature is called the Boyle Temperature, TB.


TB can be derived from the van der Waals equation of state for a real gas,



p = [RT/(V(m)-b)]-a/V(m)^2



In order to do that, we express the van der Waals equation as



p = [RT/V(m)]{1/[1-b/V(m)] - a/RTV(m)}



So long as b/Vm < 1, we can expand the first term inside the brackets according to



(1 – x)-1 = 1 + x + x2 + ...



which gives



p = [RT/V(m)]{1 + (b - a/RT)/V(m) + ...}



Obviously the virial coefficient B is given by



B = b - a/RT



At the Boyle Temperature B = 0, and so



T(B) = a/bR = 27T(c)/8

(Tc = 8a/27Rb is the critical temperature of the gas. For a derivation of the relationship of the critical constants to the van der Waals constants, see P.W. Atkins, Physical Chemistry, 3d Ed. page 31.)


8 years ago
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