91影视

An ideal diatomic gas with molecular rotation but without any molecular oscillation loses a certain amount of energy as heat.

We can write heat energy in terms of specific heat at constant volume and at constant pressure. Equating and inserting specific heats in terms of R, we can get the relation between the temperature changes in both cases. Then, using its relationship with change in the internal energy, we can find the process in which the resulting decrease in the internal energy is greater.

Formulae:

Energy transferred as heat to the body at constant pressure,$Q=n{C}_p螖T$ 鈥�(颈)

Energy transferred as heat to the body at constant volume,$Q=n{C}_v螖T$鈥�(颈颈)

Value of specific heat at constant pressure,$C_p=\frac{7}{2}R$ 鈥�(颈颈颈)

Value of specific heat at constant volume,$C_v=\frac{5}{2}R$ 鈥�(颈惫)

Change in internal energy at constant process$螖E=nC螖T$,. 鈥�(v)

Using the value of equation (iii) in equation (i), we can get the heat at constant pressure for diatomic gas as follows:

$Q=n\frac{7}{2}R螖T_p$ 鈥�(惫颈)

Using the value of equation (iv) in equation (ii), we can get the heat at constant volume for diatomic gas as follows:

$Q=n\frac{5}{2}R螖T_v$ 鈥�(惫颈颈)

The heat released at constant pressure and constant volume process is thesame.

From equation (vi) and (vii), we can get the temperature relation as follows:

$\begin{array}{c}n\frac{5}{2}R螖T_V=n\frac{7}{2}R螖T_p\\ 5螖T_V=7螖T_p\\ 螖T_v>螖T_p\end{array}$ 鈥�(惫颈颈颈)

As we know from equation (v), the change in internal energy is proportional to the change in temperature.

For constant pressure, the change in internal energy is given as:$螖E_{\mathrm{int},p}鈭�螖T_p$.

For constant volume, the change in internal energy is given as:$螖E_{\mathrm{int},v}鈭�螖T_v$.

According to equation (viii) and the above values, we can write

$螖E_{\mathrm{int},v}>螖E_{\mathrm{int},p}$

Hence,the change in the internal energy of the ideal diatomic gas is greater if the heat loss is at constant volume.