By applying a pressure of $200atm$, 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?

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

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

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 $P=200atm,$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 :

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

localid="1651133444001" $$\begin{gathered} W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}\left(-2.016\times {10}^{12}V+2.016\times {10}^9\right)dV \\ W={\left[\left(-2.016\times {10}^{12}\right)\frac{V^2}{2}+\left(2.016\times {10}^9\right)V\right]}_{{10}^{-3}}^{0.99\times {10}^{-3}} \end{gathered}$$

Substitute localid="1651133451713" $1litre$for V to solve A and B

localid="1651133447933" $$\begin{gathered} A=-\frac{19900}{1\times {10}^{-3}}.\frac{101325{\mathrm{Pa}}_{\mathrm{a}}}{1\mathrm{amm}}\mathrm{atm}.{\mathrm{m}}^3 \\ \mathrm{A}=-2.016\times {10}^{12}\mathrm{atm}.{\mathrm{m}}^{-3} \end{gathered}$$

And similarly, for B, localid="1651133455574" $B=2.016\times {10}^9\mathrm{atm}$

Substitute AV+B for P in equation (6)

localid="1651133459614" $W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}(AV+B)dV\dots \dots \left(7\right)$

Substitute localid="1651133463066" $2.016\times {10}^9\mathrm{atm}\text{for}B\text{and}-2.016\times {10}^{12}{\mathrm{atmm}}^{-3}\text{forAinequation}\left(7\right)$

localid="1651133466748" $$\begin{gathered} W=-{\int }_{{10}^{-3}}^{0.99\times {10}^{-3}}\left(-2.016\times {10}^{12}V+2.016\times {10}^9\right)dV \\ W={\left[\left(-2.016\times {10}^{12}\right)\frac{V^2}{2}+\left(2.016\times {10}^9\right)V\right]}_{{10}^{-3}}^{0.99\times {10}^{-3}} \\ W=100.8\mathrm{J} \end{gathered}$$

If constant pressure of localid="1651133470929" $200atm=2.0265\times {10}^7\mathrm{Pa}$is considered,

Work done will be localid="1651133474496" $W=-P\Delta V=-P\left(V_f-V_i\right)\dots ..\text{(8)}$

Substitute localid="1651133477862" $0.99V_o\text{for}{V}_f\text{and}{V}_o\text{for}{V}_i\text{inequation}\left(8\right)$

localid="1651133481243" $W=-P\left(0.99V_o-V_o\right)\dots ..\left(9\right)$

Again substitute localid="1651133484260" $-2.0265\times {10}^7\text{forPand}1\times {10}^{-3}\text{for}{V}_o\text{inequation}$

localid="1651133487776" $$\begin{gathered} W=-2.0265\times {10}^7\times \left(-0.01V_o\right) \\ W=2.0265\times {10}^7\times \left(-0.01\times {10}^{-3}\right) \\ W=202.65\mathrm{J} \end{gathered}$$

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 $W=100.8\mathrm{J}.$

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