Question

When a liquid is heated its density generally:
A. Decreases
B. Increases
C. Does not change
D. Decrease or increase depending upon the pressure to which it is subjected.

Answer 1

Hint: Concept of density and effect of heat on it. Basically the change in there is change in densities of liquid on being heated.

Complete step by step answer:
$ \to $Density of an object is generally how concentrated the atoms are in given volume. It is defined as the mass per unit volume.
$ \to $Density$ = \dfrac{{Mass}}{{Volume}}$.
$ \to $The materials made of atoms tightly packed in it will have higher density as solid have higher density than liquids which in turn have higher density than gases.
$ \to $Effect of heat on density of liquids
(1). On heating, the temperature of molecules in liquid increases which in turn increases their kinetic energy.
The molecules bump into each other and spread out. Hence, the volume increases.
As density $ = \dfrac{{Mass}}{{Volume}}$
And mass remains the same here but volume increases due to increase in temperature so, density of liquids decreases with increase in temperature that is on heating.
$ \to $Let ${V_0}$and $V$ be the volume of liquid at temperature $T$ and $T + \Delta T$ respectively. If $r$ is the coefficient of cubical expansion,
Then
$V = {V_0}\left( {1 + r\Delta T} \right)$…. (i)
Let $\rho $and ${\rho _0}$ be the densities of liquid at temperature $T + \Delta T$ and $T$ respectively and $M$ be its mass then
${\rho _0} = \dfrac{M}{{{V_0}}}$ and $\rho = \dfrac{M}{V}$
$ \Rightarrow {V_0} = \dfrac{M}{{{\rho _0}}}$ and $V = \dfrac{M}{\rho }$ …. (II)
Put (ii) and (i), we get
$
  \dfrac{M}{\rho } = \dfrac{M}{{{\rho _0}}}\left( {1 + r\Delta T} \right) \\
  \rho = \dfrac{{{\rho _0}}}{{\left( {1 + r\Delta T} \right)}} \\
  \rho = {\rho _0}{\left( {1 + r\Delta T} \right)^{ - 1}} \\
 $
Using binomial,
$\rho = {\rho _0}\left( {1 - r\Delta T} \right)$
With increase in temperature,
$\Delta T \to + ve \Rightarrow \rho < {\rho _0}$
So, density decreases

So, the correct answer is “Option A”.

Additional Information:
$ \to $The above equation is valid for solids also
$ \to $$r$is of the order of ${10^{ - 3}}$
$ \to $The density of water is maximum at $4^\circ C$
$ \to $For water, density increases for $0$to$4^\circ C$. So, $r$ is negative from $0$ to $4^\circ C$
$ \to $For $4^\circ C$ and higher temperatures, $r$ is positive.

Note:
In the binomial expansion, we have neglected the higher power terms as for small change in temperature, the higher powers will be further small which will have negligible effect on change in densities.