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Question
Derive the values of critical constants in terms of van der Waals constants.
Solution
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
`("P" + "an"^2/"V"^2) ("V" - "nb")` = nRT for 1 mole
From this equation, the values of critical constant PCVC and TC arc derived in terms of a and b the Van der Waals constants.
`("P" + "an"^2/"V"^2) ("V" - "b")` = RT ..........(1)
On expanding the equestion (1)
`"PV" + "a"/"V" - "Pb" - "ab"/"V"^2 - "RT"` = 0 .........(2)
Multiplying eqestion (2) by `"V"^2/"P"`,
`"V"^2/"P"("PV" + "a"/"V" - "Pb" - "ab"/"V"^2 - "RT")` = 0
`"V"^3 + "aV"/"P" - "bV"^2 - "ab"/"P" - "RRV"^2/"P"` = 0 ........(3)
equation (3) is rearranged in the powers of V
`"V"^3 - ["RT"/"P" + "b"] "V"^2 + "aV"/"P" - "ab"/"P"` = 0 .........(4)
The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume VC.
i.e. V = VC.
V – VC = 0 ……….(5)
(V – VC)3 = 0 ………(6)
(V3 – 3VCV2 + 3VC3V – VC3 = 0 ………(7)
As equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),
– 3VCV2 = `-["RT"_"C"/"P"_"C" + "b"]"V"^2` .......(8)
3VC = `"b" + "RT"_"C"/"P"_"C"` ........(9)
3VC2 = `"a"/"P"_"C"` ........(10)
VC3 = `"ab"/"P"_"C"` .....(11)
Divide equation (11) by (10)
`"V"_"C"^3/(3"V"_"C"^2) = ("ab"/"P"_"C")/("a"/"P"_"C")`
`"V"_"C"/3` = b
∴ VC = 3b ......(12)
When equation (12) is substituted in (10)
3VC2 = `"a"/"P"_"C"`
PC = `"a"/"3V"_"C"^2 = "a"/(3(3"b")^2) = "a"/(3 xx 9"b"^2) = "a"/(27"b"^2)`
∴ PC = `"a"/(27"b"^2)` ......(13)
substituting the values of VC and PC in equation (9)
3VC = `"b" + "RT"_"C"/"P"_"C"`
`3 xx 3"b" = "b" + "RT"_"C"/("a"/(27"b"^2))`
`9"b" - "b" ="RT"_"C"/"a" xx 27"b"^2`
8b = `("T"_"C"."R"27"b"^2)/"a"`
∴ TC = `(8"ab")/(27"Rb"^2) = (8"a")/(27"RB")`
TC = `(8"a")/(27"RB")` ....... (14)
Critical constant a and b can be calculated using Van der Waals Constant as follows:
a = 3VC2PC
b = `"V"_"C"/3` .......(15)
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