91Ó°ÊÓ

The table containing four situations for an ideal gas is given.

We can find the change in the internal energy using the first law of thermodynamics. Then using the relation between change in internal energy and change in temperature, we can rank the four situations in terms of temperature change of the gas.

Formulae:

According to first law of thermodynamics, the change in internal energy is given by

$\mathrm{Δ}E=Q−W$…(¾±)

The change in internal energy at constant volume, $\mathrm{Δ}E=m{C}_V\mathrm{Δ}T$ …(¾±¾±)

In situation a, the change in internal energy using the given values in equation (i) is as follows:

$\begin{array}{c}\mathrm{Δ}E=−50\text{ â¶Ä‹J}−(−50\text{ J})\\ =0\text{ J}\end{array}$

Thus, the change in temperature using the above value in equation (ii) is as follows:

$\mathrm{Δ}T=0$

In situation b, the change in internal energy using the given values in equation (i) is as follows:

$\begin{array}{c}\mathrm{Δ}E=35\text{ J}−(+35\text{ J})\\ =0\text{ J}\end{array}$

Thus, the change in temperature using the above value in equation (ii) is as follows:

$\mathrm{Δ}T=0$

In situation c, the change in internal energy using the given values in equation (i) is as follows:

$\begin{array}{c}\mathrm{Δ}E=−15\text{ J}−(−40\text{ J})\\ =25\text{ J}\end{array}$

Thus, the change in temperature using the above value in equation (ii) is as follows:

$\mathrm{Δ}T\text{>}0$

In situation d, the change in internal energy using the given values in equation (i) is as follows:

$\begin{array}{c}\mathrm{Δ}E=20\text{ J}−\left(40\text{ J}\right)\\ =−20\text{ J}\end{array}$

Thus, the change in temperature using the above value in equation (ii) is as follows:

$\mathrm{Δ}T\text{ \&±ô³Ù;}0$

Therefore, the ranking of the four situations in terms of the temperature change of the gas is.

$c>a=b>d$