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Chapter 3

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.

Step-by-Step

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}