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## Q. 3.2

For liquid acetone at 20°C and 1 bar,

$β = 1.487 \times ​10^{−3 }° C^{−1} κ = 62 \times 10^{−6} bar^{ −1} V = 1.287 cm^{3} ⋅g^{−1}$

For acetone, find:

(a) The value of (∂ P / ∂ T) V at 20°C and 1 bar.

(b) The pressure after heating at constant V from 20°C and 1 bar to 30°C.

(c) The volume change when T and P go from 20°C and 1 bar to 0°C and 10 bar.

## Verified Solution

(a) The derivative $( ∂ P / ∂ T )_{V}$ is determined by application of Eq. (3.5) to the case for which V is constant and dV = 0:

$\frac{dV}{V} =\beta dT – k dP$              (3.5)

β dT − κ dP = 0 ( const V )

or

$\left(\frac{∂ P}{∂ T} \right) _{V} = \frac{\beta }{K} = \frac{1.487 \times 10^{-3} }{62 \times 10^{-6} }= 24 bar ⋅ °C^{-1}$

(b) If β and κ are assumed constant in the 10°C temperature interval, then for constant volume Eq. (3.6) can be written:

$\frac{V_{2} }{V_{1} }= \beta (T_{2} – T_{1} ) – k (P_{2 } – P_{1} )$       (3.6)

$p_{2} = p_{1} + \frac{\beta }{k} \left(T_{2} – T_{1} \right) = 1 bar + 24 bar . °C^{−1} × 10°C = 241 bar$

(c) Direct substitution into Eq. (3.6) gives:

In $\frac{V_{2} }{V_{1} } = ( 1.487 \times 10^{−3} ) ( −20 ) − ( 62 \times 10^{−6} ) ( 9 ) = −0.0303$

$\frac{V_{2} }{V_{1} } = 0.9702 and V_{2} = ( 0.9702 ) ( 1.287 ) = 1.249 cm^{ 3}⋅g^{−1}$

Then,

$ΔV = ​V_{2} – V_{1} = 1.249 − 1.287 = −0.038 cm^{ 3} ⋅g^{ −1}$