Question

The ratio of work done by an ideal monatomic gas to the heat supplied to it in an isobaric process is:
A. $\dfrac{2}{3}$
B. $\dfrac{3}{2}$
C. $\dfrac{3}{5}$
D. $\dfrac{2}{5}$

Answer 1

We have an ideal monatomic gas in an isobaric system (constant pressure). We need to find the ratio of work done and heat supplied to the system. From the first law of thermodynamics we have the equation that relates work done, heat and internal energy. By finding the internal energy and heat change we will get the work done from the first law equation. Thus we can find the required ratio.

Formula used:
$\Delta Q=\Delta U+\Delta W$
$\Delta U=n{{C}_{v}}\Delta T$
${{C}_{v}}=\dfrac{f}{2}R$
$\Delta Q=n{{C}_{p}}\Delta T$
${{C}_{p}}=\left( \dfrac{f}{2}+1 \right)R$

Complete answer:
In the question we are given an isobaric system.
We are asked to find the ratio of work done by an ideal monatomic gas to the heat supplied, i.e.
$\dfrac{\Delta W}{\Delta Q}$, where ‘$\Delta W$’ is the work done and ‘$\Delta Q$’ is the heat supplied.
From the first law of thermodynamics, we have the equation
$\Delta Q=\Delta U+\Delta W$, where ‘$\Delta U$’ is the change in the internal energy.
We know that the change in internal energy is given by the equation,
$\Delta U=n{{C}_{v}}\Delta T$, were ‘n’ is the number of moles, ‘${{C}_{v}}$’ molar specific heat at constant volume and ‘$\Delta T$’ is the change in temperature.
We also know that the molar specific heat at constant volume is,
${{C}_{v}}=\dfrac{f}{2}R$, where ‘f’ is the degree of freedom and ‘R’ is the ideal gas constant.
In the question we are given a monatomic gas. We know that for a monatomic gas the degree of freedom is 3, i.e.
$f=3$
Therefore we get the equation for molar specific heat at constant volume as,
$\Rightarrow {{C}_{v}}=\dfrac{3}{2}R$
By substituting this in the equation for change in internal energy, we get
$\Rightarrow \Delta U=n\times \dfrac{3}{2}R\Delta T$
$\Rightarrow \Delta U=\dfrac{3}{2}nR\Delta T$
We know that for an isobaric process, the change in heat energy is given by the equation,
$\Delta Q=n{{C}_{p}}\Delta T$, were ‘n’ is the number of moles, ‘${{C}_{p}}$’ is the molar specific heat at constant pressure and ‘$\Delta T$’ is the change in temperature.
We also know that the molar specific heat at constant pressure is given by the equation,
${{C}_{p}}=\left( \dfrac{f}{2}+1 \right)R$
Therefore for the given monatomic gas, we get
$\Rightarrow {{C}_{p}}=\left( \dfrac{3}{2}+1 \right)R$
$\Rightarrow {{C}_{p}}=\dfrac{5}{2}R$
By substituting this in the equation for change in heat energy, we get
$\Rightarrow \Delta Q=n\times \dfrac{5}{2}R\Delta T$
$\Rightarrow \Delta Q=\dfrac{5}{2}nR\Delta T$
Now by substituting the value of heat energy and internal energy in the equation of first law of thermodynamics, we get
$\Rightarrow \dfrac{5}{2}nR\Delta T=\dfrac{3}{2}nR\Delta T+\Delta W$
By solving this we get the work done as,
$\Rightarrow \Delta W=\dfrac{5}{2}nR\Delta T-\dfrac{3}{2}nR\Delta T$
Now that we have the value for work done and the heat supplied, we can find its ratio,
$\Rightarrow \dfrac{\Delta W}{\Delta Q}=\dfrac{nR\Delta T}{\dfrac{5}{2}nR\Delta T}$
$\Rightarrow \dfrac{\Delta W}{\Delta Q}=\dfrac{2nR\Delta T}{5nR\Delta T}$
$\Rightarrow \dfrac{\Delta W}{\Delta Q}=\dfrac{2}{5}$
Therefore we get the ratio of the work done by an ideal monatomic gas to the heat supplied in an isobaric process as $\dfrac{2}{5}$.

Hence the correct answer is option D.

Note:
The first law of thermodynamics simply says that heat is a form of energy. All the thermodynamics processes are subject to the principle of conservation of energy. Or we can also say that energy cannot be created and destroyed, we can only convert energy from one form to another.
The mathematical expression for this law is given as,
$\Delta U=Q-W$, where ‘$\Delta U$’ is the change in internal energy, ‘Q’ is the heat supplied and ‘W’ is the work done by the system.