Question

As a result of isobaric heating $\Delta T = 72K$, one mole of a certain ideal gas obtains an amount of heat $Q = 1.60kJ$. If the value of $\gamma$ is $\dfrac{{(10 + x)}}{{10}}$. Find $x$.

Answer 1

Hint: Given that the process is isobaric, so this means that the pressure is constant. If a gas expands at a constant pressure, then this process is known as isothermal expansion. In an isobaric process work done is proportional to volume and it is reversible.

Complete step by step solution:
Step I: In an isobaric process, the work done is given by
$W = \int {P.dV} $
$W = P\int {\Delta V} $

Step II:
But according to ideal gas law,
$P\Delta V = nR\Delta T$
$W = nR\Delta T$---(ii)
$R$ is gas constant and its value is $8.314$
For one mole of ideal gas, $n = 1$
Substitute the given values in equation (ii),
$W = 1 \times 8.314 \times 72$
$W = 598.60J$
Or $W \approx 600J = 0.6kJ$

Step III: Now according to the First Law of thermodynamics, the energy can neither be created nor destroyed. It can be converted from one form to another. For first law of thermodynamics,
$\Delta U = Q - W$
Where $\Delta U$ is the change in internal energy of the system
$Q$ is the energy or heat supplied
$W$ is the amount of work done

Step IV: Substitute the given values and find the value of internal energy of the system
$\Delta U = 1.6 - 0.6$
$\Delta U = 1kJ$

Step V: In case of isobaric processes, the internal energy of the system is given by
$\Delta U = n{c_v}\Delta T$---(i)
Where ${c_v}$ is the heat capacity of the substance
$\Delta T$ is the change in temperature
And the amount of heat of the system is given by
$Q = n{c_p}\Delta T$---(ii)

Step VI:
Dividing equation (i) and (ii),
$\dfrac{Q}{{\Delta U}} = \dfrac{{n{c_p}\Delta T}}{{n{c_v}\Delta T}}$
$\dfrac{Q}{{\Delta U}} = \dfrac{{{c_p}}}{{{c_v}}}$----(iii)

Step VII: Specific heat ratio in the isobaric process is given by gamma $\gamma $. Its formula is
$\gamma = \dfrac{{{c_p}}}{{{c_v}}}$---(iv)
Comparing (iii) and (iv)
$\gamma = \dfrac{Q}{{\Delta U}}$
$\gamma = \dfrac{{1.6}}{1}$
$\gamma = 1.6$

So the value of x is $\gamma = 1.6.$

Note: It is important to note that in an isobaric process, the volume of the system is allowed to expand or contract. It is to be done in such a way that it neutralises any pressure change. In an isobaric process, work is done and due to transfer of heat, there is change in internal energy.