91Ó°ÊÓ

P-V diagram into regions 1 and 2 for the gas represents the initial state of the gas and isotherm through the dot of the graph.

Rising temperature and changes in state or phase from solid to liquid and liquid to gas cause a rise in internal energy. We can use the change in internal energy formula, which gives the relation between the change in internal energy and change in temperature to predict the change in internal energy in given cases.

Formulae:

Change in internal energy at constant volume,

$\mathrm{Δ}E=n{C}_V\mathrm{Δ}T$ …(¾±)

Change in internal energy when the gas moves up along the isotherm, i.e., change in temperature is zero. That is the change in temperature is given as follows:

$\mathrm{Δ}T=0$

So, according totheequation (i), the change in internal energy is given as follows:

$\mathrm{Δ}{E}_{int}=0$

Therefore, the change in internal energy is zero for the case of gas moving up the isotherm line.

Similarly, as gas moves down along the isotherm but process is isothermal, i.e., total change in temperature is zero. That is the change in temperature is given as follows:

$\mathrm{Δ}T=0$

So, according totheequation (i), the change in internal energy is given as follows:

$\mathrm{Δ}{E}_{int}=0$

Therefore, the change in internal energy is zero for the case of gas moving down the isotherm line.

As gas moves anywhere in region 1, the change in temperature is less than zero.

From the graph, we can say that in region 1, the final temperature is less than the initial temperature as: $T_f<T_i,$that is, the change in temperature is given as follows: $\mathrm{Δ}T<0$.

So, according totheequation (i), the change in internal energy is given as follows:

$\mathrm{Δ}{E}_{int}<0$

Therefore, the change in internal energy is negative for the case of gas moving anywhere in region 1.

As gas moves anywhere in region 2, the change in temperature is greater than zero, i.e., there is an increase in temperature.

From the graph, we can say that in region 2, the final temperature is greater than the initial temperature as: $T_f>T_i$, that is, the change in temperature is given as follows: $\mathrm{Δ}T>0$

So, according totheequation (i), the change in internal energy is given as follows:

$\mathrm{Δ}{E}_{int}>0$

Therefore, the change in internal energy is positive for the case of gas moving anywhere in region 1.