Question

A metal bar of mass 1.5 kg is heated at atmospheric pressure. Its temperature has increased from $30^\circ$ to $60^\circ$. Then the work done in the process is:
(Volume expansion coefficient=$5\times {{10}^{-5}}{}^\circ {{C}^{-1}}$
Density of metal=$9\times 10^3 kg/m^3$
Atmospheric pressure= $10^5$ Pa)
$\begin{aligned} & A.25\times {{10}^{-3}}J \\\ & B.2.5\times {{10}^{-3}}J \\\ & C.12.5\times {{10}^{-3}}J \\\ & D.1.25\times {{10}^{-3}}J \\\ \end{aligned}$

Answer 1

Hint: We have to know the thermodynamic formula for work done on a system. Also, we will use the mathematical expression for volume expansion coefficient. Then, simply using the given data, the answer can be found.

Formula used:
$dW=p dV$
$\beta =\dfrac{1}{V}.{{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}$

Complete step-by-step solution:
Let, $\rho =9\times {{10}^{3}}kg/{{m}^{3}}$, which is the density of the metal. And $m=1.5$ kg be its mass. The work done in a thermodynamic system is given by, $dW=p.dV$. Here p is pressure and ‘dV’ is the change in volume. But the question only says about change in temperature but not in volume. So, we need to modify the formula.
Volume is a function of temperature and pressure. So,
$V=V(p,T)$.
Taking differential, we obtain,
$dV={{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}.dT+{{\left( \dfrac{\delta V}{\delta p} \right)}_{T}}.dp$
Here, ${{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}$ is the rate of change of volume with temperature at constant pressure and ${{\left( \dfrac{\delta V}{\delta p} \right)}_{T}}$ is the rate of change of volume with pressure at a constant temperature. $dT=(60-30)=30$ is the change in temperature.
In the given problem, pressure is constant. So, $dp=0$. Putting this in the above equation, we get
$dV={{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}.dT$
Now, volume expansion coefficient is given by $\beta$ .
$\beta =\dfrac{1}{V}.{{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}$
$\Rightarrow \left(\dfrac{\delta V}{\delta T}\right)_p=\beta .V=\beta .\dfrac{m}{\rho}$
Therefore,
$\begin{aligned} & dW=p.dV=p.{{\left( \dfrac{\delta V}{\delta T} \right)}_{p}}.dT=p\beta \dfrac{m}{\rho }.dT \\\ & dW={{10}^{5}}\times 5\times {{10}^{-5}}\times \dfrac{1.5}{9\times {{10}^{3}}}\times 30=25\times {{10}^{-3}}J \\\ \end{aligned}$
So, option A is the correct answer.

Additional information:
In physics, when work is done by a system, it is positive work. And when work is done against the system, it is negative work. However, in chemistry the opposite is believed.

Note: Remember the following things,
1. In this problem, dT only means the difference in temperature. Its value is the same for Kelvin and °C scale.
2. Don’t try to calculate work by first law of thermodynamics here. Because we don’t know the internal energy.
3. If the work is done in constant temperature, put $dT=0$ in the expression of ‘dV’ and make $dp \neq 0$ and use corresponding given data.