Question

In an isobaric process, the correct ratio is:
(A) $\Delta Q:\Delta W = 1:1$
(B) $\Delta Q:\Delta W = \gamma :\gamma - 1$
(C) $\Delta Q:\Delta W = \gamma - 1:\gamma $
(D) $\Delta Q:\Delta W = \gamma :1$

Answer 1

Hint: The amount of heat added at constant pressure depends on the specific heat constant at constant pressure. And the work done is related to the gas constant. Both the specific heat at constant pressure and gas constant can be related. And the expression for gamma ratio is also included.

Complete step by step solution:
As the name suggests, the isobaric process is which the pressure will be constant. The expression for the heat added in the process is given as,
$\Delta Q = \Delta U + \Delta W...........\left( 1 \right)$
Where, $\Delta U$ is the change in internal energy and $\Delta W$ is the work done.
The expression for the heat added in constant pressure is given as,
$\Delta Q = n{C_p}\Delta T............\left( 2 \right)$
Where, $n$ is the number of moles, ${C_p}$ is the specific heat at constant pressure and $\Delta T$ is the change in temperature.
The constant volume molar heat capacity $\left( {{C_V}} \right)$ is independent of temperature and the expression for the change in internal energy is given as,
$\Delta U = n{C_V}\Delta T...........\left( 3 \right)$
Where, $n$ is the number of moles and $\Delta T$ is the change in temperature.
Hence the pressure is constant; there will be no force exerting. And the expression for work done will be
$
  \Delta W = P\Delta V \\
  \Delta W = nR\Delta T............\left( 4 \right) \\
$
Where, $R$ is the gas constant.
This expression is obtained from the ideal gas equation.
We want to find the ratio of $\Delta Q$ and $\Delta W$ . Therefore dividing equation $\left( 4 \right)$ from equation $\left( 2 \right)$ gives,
$
  \dfrac{{\Delta Q}}{{\Delta W}} = \dfrac{{n{C_p}\Delta T}}{{nR\Delta T}} \\
  \dfrac{{\Delta Q}}{{\Delta W}} = \dfrac{{{C_p}}}{R}..........\left( 5 \right) \\
 $
The relation connecting specific heat constant at constant pressure $\left( {{C_p}} \right)$ and specific heat constant at constant volume $\left( {{C_V}} \right)$ is
$
  {C_p} - {C_V} = R \\
  \dfrac{{{C_p} - {C_V}}}{{{C_p}}} = \dfrac{R}{{{C_p}}} \\
  1 - \dfrac{{{C_V}}}{{{C_p}}} = \dfrac{R}{{{C_p}}} \\
 $
And the ratio of $\left( {{C_p}} \right)$ and $\left( {{C_V}} \right)$ is called the specific heat ratio or gamma ratio. This is expressed as
$\dfrac{{{C_P}}}{{{C_V}}} = \gamma $
Substituting this in the above equation gives,
$
  1 - \dfrac{1}{\gamma } = \dfrac{R}{{{C_p}}} \\
  \dfrac{{\gamma - 1}}{\gamma } = \dfrac{R}{{{C_p}}} \\
  {C_p} = \dfrac{{\gamma R}}{{\gamma - 1}} \\
$
Substitute this in equation $\left( 5 \right)$
$
  \dfrac{{\Delta Q}}{{\Delta W}} = \dfrac{{\gamma R}}{{R\left( {\gamma - 1} \right)}} \\
  \dfrac{{\Delta Q}}{{\Delta W}} = \dfrac{\gamma }{{\left( {\gamma - 1} \right)}} \\
 $
Therefore the ratio is $\Delta Q:\Delta W = \gamma :\gamma - 1$
The answer is option B.

Note: The constant volume molar heat capacity $\left( {{C_V}} \right)$ is independent of temperature and it has a significant role in change in internal energy. Constant pressure and constant pressure processes can bring changes in temperature.