A cylinder fitted with a piston contains liquid methanol at 20C20^{\circ} \mathrm{C}, 100 kPa, and volume 10 L. The piston is moved, compressing the methanol to 20 MPa at constant temperature. Calculate the work required for this process. The isothermal compressibility of liquid methanol at 20C20^{\circ} \mathrm{C} is 1.22 $\times$10−9 m2/N.



Step 1

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T1=20°C=293.2KT_1 = 20\text{\textdegree} C = 293.2\text{K}

V1=10L=10103m3V_1 = 10 \text{L} = 10\cdot 10^{-3}\text{m$^3$}

P1=100kPaP_1 = 100\text{kPa}

P2=20000kPaP_2 = 20000\text{kPa}

ρ=787kg/m3\rho = 787 \text{kg/m$^3$}

βT=1.22109m2/N\beta_T = 1.22\cdot 10^{-9} \text{m$^2$/N}

Isothermal compressibility is defined by:

βT=1V(VP)T\beta_T = - \frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T

Net work is w=12PdVw = \int_{1}^{2} PdV or better =12P(VP)TdPT=\int_{1}^{2}P\left(\frac{\partial V}{\partial P}\right)_TdP_T because we can change it to

w=12P(VβT)dPTw = \int_{1}^{2}P(-V\beta_T)dP_T (since (VP)T=VβT\left(\frac{\partial V}{\partial P}\right)_T = -V\beta_T)

w=VβT2(P22P12)w = \frac{-V\beta_T}{2}(P_2^2 - P_1^2) (we have all the values)

w=1010312202(2020.12)w = \frac{-10\cdot10^{-3}\cdot1220}{2}(20^2 - 0.1^2)

w=2.44kJ\boxed{w = 2.44 \text{kJ}}

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