Question 225: Properties of Matter

Question

A certain gas is isothermally compressed to $\left(\frac{1}{3}ight)^{\mathrm{rd}}$ of its initial volume $\left(V_{\mathrm{o}}=3ight.$ litre) by applying required pressure. If the bulk modulus of the gas is $3 	imes 10^5 \mathrm{~N} / \mathrm{m}^2$, the magnitude of work done on the gas is $\_\_\_\_$ J.

Accepted Answer

The Bulk Modulus of a material is defined as the ratio of the infinitesimal pressure increase (dP) to the resulting relative decrease of the volume:
$$ B=-\frac{d P}{d V / V}=-\frac{V d P}{d V} $$
$$\Rightarrow $$ $$ \mathrm{dP}=-\mathrm{B} \frac{\mathrm{dV}}{\mathrm{~V}} . $$
The work done on a gas during a volume change is
$$ W=-\int\limits_{V_i}^{V_f} P d V $$
$$\Rightarrow $$ $W=-\int\limits_{V_i}^{V_f} B \ln \left(\frac{V_i}{V}ight) d V$
$$\Rightarrow $$ $$ \mathrm{W}=\int \Delta \mathrm{P} \mathrm{dV} $$
However, for a gas where B is constant during isothermal compression, the work done is:
$\mathrm{W}=\mathrm{B} \Delta \mathrm{V}$ for small changes or $\mathrm{W}=\int \mathrm{P} \mathrm{dV}$
The initial volume is $V_i=3$ litres $=3 	imes 10^{-3} \mathrm{~m}^3$
And the final volume is $V_f=13 	imes 3=1$ litre $=1 	imes 10^{-3} \mathrm{~m}^3$
So, the change in volume is;
$$ \Delta V=V_i-V_f=2 	imes 10^{-3} \mathrm{~m}^3 $$
The bulk modulus is $\mathrm{B}=3 	imes 10^5 \mathrm{~N} / \mathrm{m}^2$
For an isothermal process in a gas, the Bulk Modulus is equal to the pressure ( $\mathrm{B}=\mathrm{P}$ ).
$$ W=\int_{V_i}^{V_f} P d V=\int_{V_i}^{V_f} B \frac{V_i}{V} d V $$
$$\Rightarrow $$ $\mathrm{W}=\mathrm{B} \mathrm{V}_{\mathrm{i}} \ln \left(\frac{\mathrm{V}_{\mathrm{i}}}{\mathrm{V}_{\mathrm{f}}}ight)$
$$\Rightarrow $$ $\mathrm{W}=\left(3 	imes 10^5ight) 	imes\left(3 	imes 10^{-3}ight) 	imes \ln \left(\frac{\mathrm{V}_0}{\mathrm{~V}_0 / 3}ight)$
$$\Rightarrow $$ $$ W=900 	imes 1.0986 \approx 988.74 \mathrm{~J} $$
Therefore, the magnitude of work done on the gas is approximately 989 J .

Properties of Matter

Integer / Numeric Answer Type