Question

An ideal gas with adiabatic exponent γ is heated at constant pressure. It absorbs q amount of heat. Fraction of heat absorbed in increasing the temperature is

Answer 1

Answer: The fraction of heat absorbed in increasing the temperature is $\frac{\gamma -1}{\gamma}$.

Explanation:

The fraction of heat supplied is used for external work.

In the given problem, An ideal gas with adiabatic exponent γ is heated at constant pressure. It absorbs q amount of heat.

The expression for the heat at constant pressure is as follows:

$Q=nC_{p}\Delta T$                                       ........ (1)

Here, C_{p} is the specific heat at constant pressure, n is the number of moles and $\Delta T$ is the change in the temperature.

The expression for the external work done is as follows:

$W=nR\Delta T$                                              ........ (2)

Here, R is the universal gas constant.

Calculate the fraction of heat absorbed in increasing the temperature by dividing the equation by (2) and (1).

$\frac{W}{q}=\frac{R}{C_{p}}$                          ......... (3)

The expression for the specific heat capacity at constant pressure.

$C_{p}=\frac{\gamma R}{\gamma -1}$

Here, $=\gamma$ is the adiabatic exponent.

Put the expression of specific heat at constant in the equation (3).

$\frac{W}{q}=\frac{\gamma -1}{\gamma}$

Therefore, the fraction of heat absorbed in increasing the temperature is $\frac{\gamma -1}{\gamma}$.

Answer 2

"An ideal gas with adiabatic exponent γ is heated at constant pressure. It absorbs q amount of heat. Fraction of heat absorbed in increasing the internal energy is  and the fraction of heat absorbed in work done is $1 - \frac {1}{\gamma}.$

Derivation explanation

Heat absorbed by the system at constant pressure:

$Q=nC_p\deltaT$

Change in internal energy:

$\deltaU=nC_v\deltaT$

$W=Q−\deltaU$

Fraction

$\frac {U}{Q} = \frac {Q-\delta U}{Q}$

$= 1- (-\frac{\delta U}{Q})$

$= 1 - \frac {1}{\gamma}$"