Question

Using the relationship for internal pressure π_T = T (∂P/∂T)_V – P, show that for a gas that obeys a truncated virial equation of state: Z = PV_m/RT = 1 + B(T)/V_m, the internal pressure may be approximated as π_T ≈ RT²(V_m)^-2∙(ΔB/ΔT). Estimate the internal pressure at 1.0 bar and also at 10.0 bar for a hypothetical real gas at 275K given

that B(T) = -28.0 cm³⋅mol^-1 at 250K and -15.6 cm³⋅mol^-1 at 300K for this gas.

Answer 1

Solution:-

The ideal gas law can be derived from the two empirical gas laws, namely, Boyle’s law,

pV = k1, at constant T, (1)

and Charles’ law,

V = k2T, at constant p, (2)

where k1 and k2 are constants.

From Eq. (1) and (2), we may assume that V is a function of p and T.

Therefore, dV (p, T) = µ∂V ∂p ¶ T dp + µ∂V ∂T ¶ p dT. (3)

Substituting from above, we get (∂V /∂p)T = −k1/p2 = −V /p and (∂V /∂T)p = k2 = V /T.

Thus, dV = −V p dP + V T dT, or dV V + dp p = dT T .

Integrating this equation,

we get ln V + ln p = ln T + c (4)

where c is the constant of integration.

Assigning (quite arbitrarily) ec = R, we get

ln V + ln p = ln T + lnR, or pV = RT. (5)

The constant R is, of course, the gas constant, whose value has been determined experimentally.

• For gases like Hydrogen and Helium we can assume that the internal pressure may be approximated as πT ≈ RT2(Vm)-2∙(ΔB/ΔT). Estimate the internal pressure at 1.0 bar and also at 10.0 bar for a hypothetical real gas at 275K given that B(T) = -28.0 cm3⋅mol-1 at 250K and -15.6 cm3⋅mol-1 at 300K for this gas.

These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small. So a/V2 is negligible even at ordinary temperatures.

Thus PV > RT. Thus Vander Waals equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation. Vander Waals equation accounts for the behaviour of real gases.

At low pressures, the gas equation can be written as,   

(P + a/v2m) (Vm) = RT

or

Z = Vm / RT = 1 – a/VmRT

Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of Vm, Z has more value at higher temperature.

At high pressures, the gas equation can be written as

P (Vm – b) = RT

Z = PVm / RT = 1 + Pb / RT

Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure. For the gases H2 and He, the above behaviour is observed even at low pressures, since for these gases, the value of ‘a’ is extremely small.  

Thus by above explaination we are in a state to prove that

Z = PVm/RT = 1 + B(T)/Vm