Question

When heat is supplied to a diatomic gas at constant pressure, then the ratio of $\Delta Q:\Delta U:\Delta W$will be
A. $5:2:3$
B. $7:2:5$
C. $5:3:2$
D. $7:5:2$

Answer 1

Diatomic molecules are molecules that are composed of only two atoms, of the same or different chemical elements. Recall the expressions for change in heat, change in internal energy and work done in the constant pressure process. For the diatomic gas, the degrees of freedom are 5. Use this and find out the specific heat and constant pressure and specific heat at constant volume.

Complete step by step answer:
We know that the process where the pressure remains the same is known as isobaric process. For the Isobaric process, Change in heat energy, $\Delta Q = nCp\Delta T$ (where $Cp$ is molar heat capacity at a constant pressure). Change in the internal energy, $\Delta U = nCv\Delta T$ (where $Cv$ is molar heat capacity at a constant pressure)

From the thermodynamics law, we have,
$Q = \Delta U + W$
Where W is work, $\Delta U$ is internal energy, and Q is heat. Work done by the closed system is defined as:
$W = \smallint pdV$
Where $\Delta$ means change over the whole process, whereas $d$denotes a differential. Since pressure is constant, this means that
$W = P\Delta V$

Applying the ideal gas law, this becomes.
$W = nR\Delta T$
With $R$ represent the gas constant and $n$ represent the number of moles.
$W = P\Delta V = nR\Delta T$
Hence,
$\Delta Q:\Delta U:\Delta W = Cp:Cv:R$
For diatomic gas, number of degrees of freedom $f = 5$
So we get,
$Cv = \dfrac{{fR}}{2} = \dfrac{{5R}}{2}$
Also,
$Cp = Cv + R \\\ \Rightarrow Cp = \dfrac{{5R}}{2} + R \\\ \Rightarrow Cp = \dfrac{{7R}}{2}$

Therefore, we have, $Cp = \dfrac{{7R}}{2}$, $Cv = \dfrac{{5R}}{2}$.
Thus, the ratio,
$\Delta Q:\Delta U:W = \dfrac{{7R}}{2}:\dfrac{{5R}}{2}:R$
$\therefore \Delta Q:\Delta U:W = 7:5:2$

So, the correct answer is option D.

Additional Information:
In thermodynamics, an isobaric process is a type of thermodynamic process in which the pressure of the system stays constant: $\Delta P = 0.$The heat transferred to the system does work, but also changes the internal energy $\left( {IS} \right)$ of the system. The articles use the physics sign convention for work, where positive work is work done by the system. Using this convention, by the first law of thermodynamics,
$Q = \Delta U + W.$

Note: When we wanted to find $Q,$sometimes we use $Q = n\left( {p\Delta \operatorname{T} } \right)$and sometimes we use $Q = n\left( {V\Delta T + p\left( {{V_f} - {V_i}} \right)} \right)$, where, n is the number of moles, V is the constant volume, p is the pressure, ${V_f}$ is the final volume and ${V_i}$ is the initial volume. They are not different things. Both of the above expressions are correct and can be used simultaneously or separately depending on the situation.