Question

An external pressure $P$ is applied on a cube at $0^\circ C$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:
A) $\dfrac{{3\alpha }}{{PK}}$
B) $3PK\alpha$
C) $\dfrac{P}{{3\alpha K}}$
D) $\dfrac{P}{{\alpha K}}$

Answer 1

Hint Let, $V$ be the initial volume of the cube and $dV$ be the corresponding decrease in volume, then, the bulk modulus of the cube is $K = \dfrac{p}{{ - \dfrac{{dV}}{V}}} = - \dfrac{{pV}}{{dV}}$ . Now, if $dT$ be the corresponding change in temperature, then, the coefficient of the volume expansion becomes, $\gamma = \dfrac{{dV}}{{VdT}}$ . Also, $\gamma = 3\alpha$ . Put this value in the formula of the coefficient of the volume expansion and simplify.
Formula used Let, $V$ be the initial volume of the cube and $dV$ be the corresponding decrease in volume, then, the bulk modulus of the cube is $K = - \dfrac{{pV}}{{dV}}$
If $dT$ be the corresponding change in temperature, then, the coefficient of the volume expansion becomes, $\gamma = \dfrac{{dV}}{{VdT}}$ and $\gamma = 3\alpha$ .

Complete step by step answer
Within the elastic limit, volume stress divided by volume strain is called the bulk modulus of elasticity.
Let, $V$ be the initial volume of the cube and $dV$ be the corresponding decrease in volume due to compression. Also, it is given that $P$ is the applied external pressure.
So, the volume strain will be $- \dfrac{{dV}}{V}$ .
Therefore, the bulk modulus of the material of the cube is $K = \dfrac{p}{{ - \dfrac{{dV}}{V}}} = - \dfrac{{pV}}{{dV}}$ .
So, $dV = - \dfrac{{pV}}{K}$
Now, the increase in volume for unit rise in temperature for a unit volume of a solid is called the coefficient of volume expansion of the material of that solid.
So, we have the initial volume of the cube, $V$ and the corresponding change in volume $dV$ .
Let, $dT$ be the corresponding change in temperature for the change in the body.
So, the coefficient of the volume expansion becomes, $\gamma = \dfrac{{dV}}{{VdT}}$
or, $dV = \gamma VdT$ $\ldots \left( 1 \right)$
Now, it is observed that the linear expansion of a body on heating is directly proportional to the initial length of the body and the rise in temperature of the body.
It is given that $\alpha$ is the coefficient of linear expansion.
Now, it can be proven that $\gamma = 3\alpha$
So, the equation $\left( 1 \right)$ can be written as
$dV = V\left( {3\alpha } \right)dT$
or, $\dfrac{{pV}}{K} = V(3\alpha )dT$
or, $dT = \dfrac{p}{{3\alpha K}}$
So, the temperature should be raised by $\dfrac{p}{{3\alpha K}}$ .

Note Compressibility is defined as the change in volume due to a unit change in pressure.
Let, $V$ be the initial volume of a body and $dV$ be the corresponding decrease in volume due to compression and also, $\Delta P$ is the applied external pressure,
Then, from the definition, compressibility is $\dfrac{{ - dV}}{{V\Delta P}}$ .
In the limit $dP \to 0$ , we have, compressibility, $- \dfrac{1}{V}\dfrac{{dV}}{{dP}} = \dfrac{1}{K}$
So, compressibility is the reciprocal of the bulk modulus.