## Related questions with answers

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature changes from $0^{\circ} \mathrm{C} \text { to } 20^{\circ} \mathrm{C}$ Determine $\Delta V^{t}, W, Q, \Delta H^{t}, \text { and } \Delta U^{t}.$ The properties for liquid carbon tetrachloride at 1 bar and $0^\circ C$ may be assumed independent of temperature: $\beta=1.2 \times 10^{-3} \mathrm{K}^{-1}, C_{P}=0.84 \mathrm{kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}, \text { and } \rho=1590 \mathrm{kg} \cdot \mathrm{m}^{-3}$

Solution

VerifiedGiven: Mass of Liquid tetrachloride is: $M=5$ $\mathrm{Kg}$

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Properties at $1$ $\mathrm{bar}$ and $T=0 \text{\textdegree} C$ is
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$\beta=1.2\times 10^{-3} \hskip 0.5em \mathrm{K^{-1}} \hskip 0.5em C_{p}=0.84 \hskip 0.5em \mathrm{KJ\cdot Kg^{-1}\cdot k^{-1} }\hskip 0.5em \rho =1590\hskip 0.5em \mathrm{kg\cdot m^{-3}} \hskip 0.5em$

Heat the system from $0\text{\textdegree} C$ to $20\text{\textdegree} C$ by mechanically reversible at constant pressure $1$ bar

Since we know from definition of expansivity $\beta$

$\beta=\dfrac{1}{V}\dfrac{dV}{dT}\bigg|_{P}$

$\int \dfrac{dV}{V}=\beta\int dT$

$\ln \dfrac{V_{2}}{V_{1}}=\beta\left( T_{2}-T_{1}\right)$

Since we know $T_{1}$, $T_{2}$ and $V_{1}=\dfrac{M}{\rho}=\dfrac{5}{1590}$ then by rearranging and substituting given value we get

$dV=V_{1}\left(\exp \left(\beta\left( T_{2}-T_{1}\right) \right)-1 \right)$

$dV=\dfrac{5}{1590}\left(\exp \left(1.2\times 10^{-3}(20-0) \right)-1 \right)=7.638\times 10^{-5} \hskip 0.5em \mathrm{m^{-3}}$

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