Question

A gas for which $\gamma$ is $\dfrac{4}{3}$ is heated at constant pressure. The percentage of heat supplied used for external work is
(A) $25\%$
(B) $75\%$
(C) $60\%$
(D) $40\%$

Answer 1

Hint
From the first law of thermodynamics we can find the relation between the work done, change in internal energy and heat. Now the head supplied for external work is given by, $\dfrac{{dw}}{{dq}}$. So by substituting the values and using the value of $\gamma$ from the equation, we get the answer.
In this solution we will be using the following formula,
$dq = du + dw$
where $dq$ is the heat supplied and has value of $n{C_p}\Delta T$
$du$ is the change in the internal energy given by $n{C_v}\Delta T$and $dw$ is the work done given by $nR\Delta T$.

Complete step by step answer
From the first law of thermodynamics the work done, heat supplied and the internal energy of a system in a constant pressure heating process are related by the formula.
$dq = du + dw$
Now the heat supplied can be written in terms of the specific heat as, $dw = n{C_p}\Delta T$ and the change in the internal energy can be written in terms of specific heat as $du = n{C_v}\Delta T$
where ${C_p}$ and ${C_v}$ are the specific heat at constant pressure and change in temperature is denoted as, $\Delta T$.
The work done by the system is given by, $dw = nR\Delta T$
So now substituting the values I the first law we get,
$dq = n{C_v}\Delta T + nR\Delta T$
Now the fraction heat supplied that is used for external work is given by, $\dfrac{{dw}}{{dq}}$
Therefore, now substituting the values we get,
$\dfrac{{dw}}{{dq}} = \dfrac{{nR\Delta T}}{{n{C_v}\Delta T + nR\Delta T}}$
Now here we can cancel the change in temperature $\Delta T$ and the number of moles $n$ from the numerator and denominator. So we get
$\dfrac{{dw}}{{dq}} = \dfrac{R}{{{C_v} + R}}$
The value of ${C_v}$ is given by, ${C_v} = \dfrac{R}{{\gamma - 1}}$
So substituting the value in the equation we get,
$\dfrac{{dw}}{{dq}} = \dfrac{R}{{\dfrac{R}{{\gamma - 1}} + R}}$
Cancelling $R$ from the numerator and the denominator, we get
$\dfrac{{dw}}{{dq}} = \dfrac{1}{{\dfrac{1}{{\gamma - 1}} + 1}}$
In the denominator for the addition of the fraction we can take LCM as $\gamma - 1$ and then move it to the numerator,
$\therefore \dfrac{{dw}}{{dq}} = \dfrac{{\gamma - 1}}{{1 + \gamma - 1}}$
From the denominator the 1 gets cancelled and we have,
$\dfrac{{dw}}{{dq}} = \dfrac{{\gamma - 1}}{\gamma }$
Now in the question we are given the value of $\gamma$ as $\dfrac{4}{3}$. So substituting the value we get,
$\dfrac{{dw}}{{dq}} = \dfrac{{\dfrac{4}{3} - 1}}{{\dfrac{4}{3}}}$
On simplifying and doing the subtraction of the fractions in the numerator we get,
$\dfrac{{dw}}{{dq}} = \dfrac{{\left( {4 - 3} \right)}}{3} \times \dfrac{3}{4}$
So the 3 gets cancelled. Hence we get
$\dfrac{{dw}}{{dq}} = \dfrac{1}{4}$
To get the percentage, we multiply it by $100\%$ on both sides which gives us,
$\dfrac{{dw}}{{dq}} \times 100\% = \dfrac{1}{4} \times 100\%$
On calculating we get,
$\dfrac{{dw}}{{dq}} \times 100\% = 25\%$
Hence the percentage of heat supplied used for external work is, $25\%$
So the correct answer is option (A).

Note
The first law of thermodynamics is used in this problem. It states that the change in internal energy of a system is equal to the net transfer of heat of the system and the net work done by the system. The equation is written in the form $du = dq - dw$.