Physical Chemistry . A certain real gas obeys the equation of van der Waals Calc

Question

Physical Chemistry

. A certain real gas obeys the equation of van der Waals Calculate the isotherms of this gas for different temperatures, using the values a = b = R = 1. With values for P0 to 0.2 and V0 to 10 (without units). Verify that at high temperatures the isotherms resemble those of an ideal gas and that Low temperatures distort and move away from ideal behavior a. b. Calculate Z for P 0 to 0.5. K-*( %P, Under For this gas, derive the expression for isothermal compressibility what conditions does the expression obtained it can be the expression for an ideal gas? c.

Explanation / Answer

a) When the pressure is not very high,the volume V will be sufficiently large,and b may be ignored in comparison.Then the van der Vaals equation becomes

(P +a/V2) (V-b) = RT

(P +a/V2) V = RT If b is ignored

PV+a/V = RT

PV = RT-a/V

As pressure increases volume V decreases,a/V increases, and therefore PV becomes smaller and smaller.This explains the dip in the isothermsof most of the gases.

if we put P= 0 and V=0 and a=R=1

Then T=1

When the pressure is considerably high,the volume V will be quite small, now it may not be possible to ignore b.But the pressure is quite high and a/V2 may become negligible in comparison with P.Then the van der Vaals equation becomes

P (V-b) = RT

PV = RT+Pb

If we put P=0.2, V=10 and b=R=1

Then

(0.2 x 10) = 1 xT+( 0.2 x 1)

2 = T +0.2

T = 1.8

At high temperatures b may also be negligible in comparison to V which is sufficiently large.Under these conditions the van der Vaals equation becomes the ideal equation PV = RT

b)

Z = PV/RT

If P =0 ,V=0 R = 1 ,T=1

Z=0

C)

Isothermal compressability factor

cg = (1/p)-1/z(dz/dp)T

As the pressure is increased the total volume occupied by the gas becomes small enough that the volume of the molecules themselves isappreciable and must be considerable. Under these conditions the expression obtained becomes ideal gas.

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