Question

An ideal gas ($\gamma = 1.4$ ) expands from $5 \times {10^{ - 3}}{m^3}$ to $25 \times {10^{ - 3}}{m^3}$ at a constant pressure of $1 \times {10^5}Pa$. The heat energy supplied during the process;
Option A: 7 J
Option B: 70 J
Option C: 700 J
Option D: 7000 J

Answer 1

Hint:The pressure is constant and so the change in heat will be equal to change in entropy. Here the relation between specific heat at constant pressure a specific heat at constant volume will be used.

Complete solution:
There are different types of thermodynamic processes. One of them is the process in which the pressure is constant; it is called the isobaric process. And one of them is the expansion process, in which the volume of the system increases.
In the question the system seems to be a combination of both the isobaric and expansion processes and so I called the isobaric expansion processes.
Since it is an expansion process the thermodynamic work done will be positive.
So, $W = P({V_2} - {V_1})$
$ \Rightarrow W = {10^5}(25 \times {10^{ - 3}} - 5 \times {10^{ - 3}})$
$ \Rightarrow W = 2000J$
For an isobaric process we know that;
$\Delta P = {P_2} - {P_1} = 0$
Also, in a process where pressure is constant the change in heat is equal to change in entropy.
Thus mathematically we can write,
$\Delta Q = \Delta H$
$ \Rightarrow \Delta Q = n{C_p}({T_2} - {T_1})$
$ \Rightarrow n{C_p} = \dfrac{{\Delta Q}}{{({T_2} - {T_1})}}$ (Equation: 1)
Also we know that, $\dfrac{{{C_p}}}{{{C_v}}} = \gamma $
$ \Rightarrow {C_v} = \dfrac{{{C_p}}}{\gamma }$ (Equation: 2)
Now let us write the first law of thermodynamics;
According to the first law of thermodynamics the heat change is equal to the sum of change in internal energy and the thermodynamic work done during the process, mathematically written as;
$\Delta Q = \Delta U + W$ (Equation: 3)
Now internal energy can be written as; $\Delta U = n{C_v}({T_2} - {T_1})$ (Equation: 4)
So from all the previous equations we get;
$n{C_p}\Delta T = n{C_v}\Delta T + P\Delta V$
$ \Rightarrow n{C_p}\Delta T = n\dfrac{{{C_p}}}{\gamma }\Delta T + 2000J$
$ \Rightarrow n\Delta T\left( {{C_p} - \dfrac{{{C_p}}}{\gamma }} \right) = 2000J$
$ \Rightarrow n{C_p}\Delta T\left( {\dfrac{{\gamma - 1}}{\gamma }} \right) = 2000J$
From equation: 1 we get;
$ \Rightarrow \left( {\dfrac{{\gamma - 1}}{\gamma }} \right)\Delta Q = 2000J$
$ \Rightarrow \Delta Q = \dfrac{{2000\gamma }}{{\gamma - 1}} = \dfrac{{2000 \times 1.4}}{{0.4}}$
$\therefore \Delta Q = 7000J$

Option D is the correct answer.

Note:
-Work done in the expansion process is positive, the final volume is greater than the initial volume.
-Work done in the compression process is negative as the final volume is less than the initial volume.
-Work done on the system is negative.
-Work done by the system is positive.