Question

Heat is supplied to a diatomic gas at constant pressure. The ratio between heat energy supplied and work done is:
($\gamma$ for diatomic gas $= \dfrac{7}{5}$)
a) 3 : 4
b) 2 : 1
c) 7 : 2
d) 2 : 5

Answer 1

Hint: The first law of thermodynamics relates the heat energy supplied to the system to the work done. The heat supplied to the system can be related to the temperature change and heat capacity of the system.
Formula used:
The first law of thermodynamics is given as
$Q = \Delta U + W{\text{ }}...{\text{(i)}}$
where Q is used to represent the heat supplied to the system, $\Delta U$ denotes the change in internal energy of the system and W represents the work done.
The change in internal energy of a thermodynamic system is given as
$\Delta U = n{C_V}\Delta T{\text{ }}...{\text{(ii)}}$
where ${C_V}$ signifies the specific heat capacity of the gas at constant pressure.

Detailed step by step solution:
We are given a diatomic gas for which we are given
$\gamma = \dfrac{7}{5}$
The heat is transferred to the system at constant pressure which means that we are dealing with heat capacity of the gas at constant pressure denoted by ${C_p}$. The expression for heat transferred to the system can be given as
$Q = n{C_p}\Delta T{\text{ }}...{\text{(iii)}}$
where n represents the no. of moles of the gas and $\Delta T$ is the change in temperature of the gas due to supplied heat energy.
The expression for the first law of thermodynamics is given by equation (i). Using equation (ii) and (iii) in equation (i), we get

$$Q = \Delta U + W \\\ \Rightarrow n{C_p}\Delta T = n{C_V}\Delta T + W \\\ \Rightarrow W = n\left( {{C_p} - {C_V}} \right)\Delta T{\text{ }}...{\text{(iv)}} \\\$$

Now we can write
$\dfrac{Q}{W} = \dfrac{{n{C_p}\Delta T}}{{n\left( {{C_p} - {C_V}} \right)\Delta T}} = \dfrac{{{C_p}}}{{{C_p} - {C_V}}}{\text{ }}...{\text{(v)}}$
Since $\gamma = \dfrac{{{C_p}}}{{{C_V}}} \Rightarrow {C_p} = \gamma {C_V}$. Using this in equation (v), we get
$\dfrac{Q}{W} = \dfrac{{\gamma {C_V}}}{{\gamma {C_V} - {C_V}}} = \dfrac{\gamma }{{\gamma - 1}}$
Using the given value of $\gamma$ for diatomic gas, we get
$\dfrac{Q}{W} = \dfrac{{\left( {\dfrac{7}{5}} \right)}}{{\left( {\dfrac{7}{5}} \right) - 1}} = \dfrac{7}{{7 - 5}} = \dfrac{7}{2}$
Hence, the correct answer is option C.

Note: We know that the first law of thermodynamics states that when a certain amount of heat is supplied to a thermodynamic system then a part of this heat energy is used to change the internal energy of the gas while rest of the energy is used in doing work.