An ideal gas is taken from a to b on the pV-diagram shown in Fig. E19.15. During this process, 700 J of heat is added and the pressure doubles. (a) How much work is done by or on the gas? Explain. (b) How does the temperature of the gas at a compare to its temperature at b? Be specific. (c) How does the internal energy of the gas at a compare to the internal energy at b? Be specific and explain.

The given data can be listed below as:

• The heat added is,$Q=700J$.
• The pressure at a is,$P_a=30kPa$.
• The volume at a and b is,$V_a=V_b=0.050m^3$ .

The work done establishes the relation between the system’s heat energy and the system’s internal energy. It can be measured in Joules.

In the process a-b, the volume is kept constant; therefore the work done by the process or the work done on or by the gas is zero. That is:

$W=0$

Here, W is the work done.

Thus, the work done is zero.

We know from the Ideal gas equation, that is:

$pV=nRT$

Here, p is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

In the above pV diagram, n, R and V are constant whereas the relation between p and T is directly proportional. That means, if the pressure increases the pressure increases too.

So, at the point b, the pressure is twice than that at the point a. That is:

$p_b=2p_a$

Here,$p_a$ is the pressure at point a, and$p_b$ is the pressure at point b.

Since the pressure doubled, the temperature will also double, that is:

$T_b=2T_a$

Here, $T_a$is the temperature at a, and $T_b$ is the temperature at b.

Therefore, the temperature at b is two times the temperature at a.

We have seen that work, $W=0$, and the heat added $Q=700J$, now according to the first law of thermodynamics the change in internal energy is given by:

$∆U=Q-W$

Substitute the values in the above equation.

$$\begin{gathered} ∆U=700J-0J \\ ∆U=700J \end{gathered}$$

Now, consider the internal energy at point a is$U_a$ and that at point b is $U_b$, then the relation is given by:

$$\begin{gathered} U_b-U_a=700J \\ {U}_b=U_a+700J \end{gathered}$$

Therefore, the change in internal energy is positive and the energy always increases as the temperature increases.

Thus, the internal energy at b is 700 J more than the internal energy at a.