Question

For four situations for an ideal gas, the table gives the energy transferred to or from the gas as heat Qand either the work W done by the gas or the work ${\mathbf{\text{W}}}_{\mathbf{o}\mathbf{n}}$done on the gas, all in joules. Rank the four situations in terms of the temperature change of the gas, most positive first.

Short answer

The ranking of four situations in terms of temperature change of the gas is$c>a=b>d$

Step-by-step solution

Stating thegiven data

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

Understanding the concept of internal energy and temperature

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

$\Delta E=Q-W$…(i)

The change in internal energy at constant volume, $\Delta E=m{C}_V\Delta T$ …(ii)

Calculation of the ranking of the situations due to temperature change

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

$\begin{array}{c}\Delta E=-50\text{\hspace{0.17em}J}-(-50\text{\hspace{0.17em}J})\\ =0\text{\hspace{0.17em}J}\end{array}$

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

$\Delta T=0$

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

$\begin{array}{c}\Delta E=35\text{\hspace{0.17em}J}-(+35\text{\hspace{0.17em}J})\\ =0\text{\hspace{0.17em}J}\end{array}$

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

$\Delta T=0$

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

$\begin{array}{c}\Delta E=-15\text{\hspace{0.17em}J}-(-40\text{\hspace{0.17em}J})\\ =25\text{\hspace{0.17em}J}\end{array}$

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

$\Delta T\text{>}0$

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

$\begin{array}{c}\Delta E=20\text{\hspace{0.17em}J}-\left(40\text{\hspace{0.17em}J}\right)\\ =-20\text{\hspace{0.17em}J}\end{array}$

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

$\Delta T\text{\hspace{0.17em}<}0$

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

$c>a=b>d$