Question

An ideal gas with adiabatic exponent $\gamma$ is heated at constant pressure. It absorbs Q amount of heat. Fraction of heat absorbed in increasing the temperature is:
A. $160\gamma$
B. $\dfrac{1}{\gamma }$
C. 1 - $\dfrac{1}{\gamma }$
D. 2$\gamma$

Answer 1

In order to solve this problem you need to know that the ratio of specific heat capacity at constant pressure to that of specific heat capacity at constant volume is $\gamma$. We also need to know that heat absorbed in increasing the temperature will be equal to the internal energy. Knowing these things and using the formula of heat will give you the right answer.

Complete answer:
It is given that an ideal gas with adiabatic exponent γ is heated at constant pressure. It absorbs Q amount of heat. We need to find the fraction of heat absorbed in increasing the temperature.
From the information above we can know that
$Q = n{C_p}\Delta T$
${C_p}$ is the heat absorbed at constant pressure $n$ is the number of moles and $\Delta T$ is the change in the temperature.
We know that heat absorbed in increasing the temperature will be equal to the internal energy.
So, change in the internal energy = $\Delta U = n{C_v}\Delta T$, ${C_p}$ is the specific heat at constant pressure.
Then the fraction of heat absorbed will be =$\dfrac{{{\text{Heat absorbed at constant pressure}}}}{{{\text{Heat absorbed at constant volume}}}}$ = $\dfrac{{\Delta U}}{Q}$
So, we do $\dfrac{{n{C_v}\Delta T}}{{n{C_p}\Delta T}} = \dfrac{{{C_v}}}{{{C_p}}}$.

So, the answer is $\dfrac{{{C_v}}}{{{C_p}}}$ but we know that $\dfrac{{{C_p}}}{{{C_v}}} = \gamma$ then $\dfrac{{{C_v}}}{{{C_p}}}$ = $\dfrac{1}{\gamma }$.

So, the correct answer is “Option B”.

Note:
In such problems you need to know the meaning of all the concepts and the meaning of respective symbols. The specific heat capacity of a substance is the heat capacity of a sample of the substance divided by the mass of the sample. Informally, it is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in its temperature. Knowing this will help you to solve further your problems.