Q. 1.32 - An Introduction to Thermal Physics

Question

By applying a pressure of $200 atm$, you can compress water to $99\%$ of its usual volume. Sketch this process (not necessarily to scale) on a $PV$ diagram, and estimate the work required to compress a liter of water by this amount. Does the result surprise you?

Step-by-Step Solution

Answer

Work done at constant pressure is $W=202.65 \mathrm{J}$ and at variable pressure work done is $W=100.8 \mathrm{J}.$

Step 1 : Given information

By applying a pressure of $200 atm$, water can be compressed to $99\%$ of its usual volume.

Step 2 : Calculation

To obtain work done on the system :$W=-{\int }_{V_t}^{v_f}{P}_{dV}\dots ...\left(1\right)$

Here P is pressure, $V_i \text{and} V_f$ are initial volume and final volume.

At a pressure of $200 atm$, the volume of the water is reduced to $99\%$ of its value at atmospheric pressure $1 atm$.

Assume that the reduction in volume is linearly proportional to the increase in pressure.

The pressure is calculated as :$P=AV+B\dots ...\text{ (2) }$

Here A and B are constant, P is pressure and V is volume.

At $P=1 \mathrm{atm}$, the volume is $V=V_o.$

And at $P=200 \mathrm{atm}$, the volume is $V=\frac{99V_o}{100}.$

Let's take the standard temperature and pressure, $P=1 \mathrm{atm}$, equation (2) can be written as by substitute $V_o \text{for} V$

$I=A{V}_o+B.\dots ..\text{ (3) }$

For $P=200 atm,$ equation (2) can be written as, by substituting $\frac{99V_o}{100} \text{for} V$

$200=A\frac{99V_o}{100}+B\dots ...\left(4\right)$

Solve equation (3) and (4) for the value of A and B

$$\begin{gathered} A=\frac{-19900}{V_o} \mathrm{atm}.{\mathrm{m}}^{-3} \\ B=19901 \mathrm{atm}. \end{gathered}$$

Now equation (1) can be written as

$P=\frac{-19900}{V_o}V+19901\dots ..\left(5\right)$

Work required to compress 1 liter of water to $99\%$ can be found by :  

$W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}PdV\dots \dots \left(6\right)$

$$\begin{gathered} W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}\left(-2.016\times {10}^{12}V+2.016\times {10}^9\right)dV \\ W={\left[\left(-2.016\times {10}^{12}\right)\frac{V^2}{2}+\left(2.016\times {10}^9\right)V\right]}_{{10}^{-3}}^{0.99\times {10}^{-3}} \end{gathered}$$

Substitute $1 litre$ for V to solve A and B

$$\begin{gathered} A=-\frac{19900}{1\times {10}^{-3}}.\frac{101325{ \mathrm{Pa}}_{\mathrm{a}}}{1\mathrm{amm}}\mathrm{atm}.{\mathrm{m}}^3 \\ \mathrm{A}=-2.016\times {10}^{12} \mathrm{atm}.{\mathrm{m}}^{-3} \end{gathered}$$

And similarly, for B, $B=2.016\times {10}^9 \mathrm{atm}$

Substitute AV+B for P  in equation (6)

$W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}(AV+B)dV\dots \dots \left(7\right)$

Substitute $2.016\times {10}^9 \mathrm{atm} \text{for} B \text{and} -2.016\times {10}^{12} {\mathrm{atmm}}^{-3} \text{for A in equation}\left(7\right)$

$$\begin{gathered} W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}\left(-2.016\times {10}^{12}V+2.016\times {10}^9\right)dV \\ W={\left[\left(-2.016\times {10}^{12}\right)\frac{V^2}{2}+\left(2.016\times {10}^9\right)V\right]}_{{10}^{-3}}^{0.99\times {10}^{-3}} \\ W=100.8 \mathrm{J} \end{gathered}$$

If constant pressure of $200 atm=2.0265\times {10}^7 \mathrm{Pa}$ is considered,

Work done will be $W=-P\Delta V=-P\left(V_f-V_i\right)\dots ..\text{ (8) }$

Substitute $0.99 V_o \text{for} V_f \text{and} V_o \text{for} V_i \text{in equation}\left(8\right)$

$W=-P\left(0.99V_o-V_o\right)\dots ..\left(9\right)$

Again substitute $-2.0265\times {10}^7 \text{for P and} 1\times {10}^{-3} \text{for }{V}_o \text{in equation}$

$$\begin{gathered} W=-2.0265\times {10}^7\times \left(-0.01V_o\right) \\ W=2.0265\times {10}^7\times \left(-0.01\times {10}^{-3}\right) \\ W=202.65 \mathrm{J} \end{gathered}$$

And the graph between pressure and volume is shown below which is followed by equation (5)

Hence, at constant pressure work done is $W=202.65 \mathrm{J}$ and at variable pressure work done is $W=100.8 \mathrm{J}.$

Step 3 : Final answer

Work done at constant pressure is $W=202.65 \mathrm{J}$ and at variable pressure work done is $W=100.8 \mathrm{J}.$