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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.
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,
In order to do that, we express the van der Waals equation as
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
Obviously the virial coefficient B is given by
At the Boyle Temperature B = 0, and so
(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.)
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