In a container with a heat-conductive wall and a piston, there is $V_{0}=1 \mathrm{~m}^{3}$ volume, $10^{5} \mathrm{~Pa}$ pressure, and $300 \mathrm{~K}$ temperature nitrogen. We perform two experiments with the gas.

a) We change the state of the gas in such a way that the ratio $\Delta p / \Delta V$ is kept at a constant value $a_{1}$. We observe that no matter whether we increase or decrease the volume of the nitrogen relative to $V_{0}$, the nitrogen always condenses.

b) In the second case, we also ensure that $\Delta p / \Delta V=a_{2}$ remains constant. Here, we observe that no matter whether we decrease or increase the volume of the gas relative to $V_{0}$, the system releases heat to its environment.

Determine the values of $a_{1}$ and $a_{2}$!
I. solution.
a) $\mathrm{A} \frac{\Delta_{p}}{\Delta_{V}}=a_{1}=$ the state condition means that during such a process, the state of the system, when plotted on a $p-V$ diagram, results in a straight line.

$$
\frac{p-p_{0}}{V-V_{0}}=a_{1}
$$

from this:

$$
p=a_{1} V+\left(-a_{1} V_{0}+p_{0}\right)
$$

on the other hand:

$$
p V=N k T
$$

thus the temperature during the process is

$$
T(V)=\frac{T_{0}}{p_{0} V_{0}} p V=\frac{T_{0}}{p_{0} V_{0}}\left[a_{1} V^{2}+\left(-a_{1} V_{0}+p_{0}\right) V\right]
$$

From the fact that this function has a maximum at $V_{0}$ (by differentiation or by completing the square), we get

$$
a_{1}=-\frac{p_{0}}{V_{0}}=-10^{5} \mathrm{~Pa} / \mathrm{m}^{3}
$$

b) During the second process

$$
\frac{p-p_{0}}{V-V_{0}}=a_{2}, \text { so } p=a_{2} V+p_{0}-a_{2} V_{0}
$$

Applying the first law of thermodynamics, the heat absorbed during the process is:

$$
Q=\Delta E+W^{\prime}=\frac{5}{2} N k\left(T-T_{0}\right)+\frac{\left(V-V_{0}\right)\left(p+p_{0}\right)}{2}
$$

where we used the fact that the heat capacity of diatomic nitrogen gas is $\frac{5}{2} N k, W^{\prime}$ is the area of the trapezoid under the $p-V$ state curve. Substituting $T=\frac{p V T_{0}}{p_{0} V_{0}} t$ and expressing $p$ in terms of $V$:

$$
Q(V)=3 a_{2} V^{2}+V\left(\frac{7}{2} p_{0}-\frac{7}{2} a_{2} V_{0}\right)-\frac{7}{2} p_{0} V_{0}+\frac{1}{2} a_{2} V_{0}^{2}
$$

The function $Q(V)$ must have a maximum at $V_{0}$, so

$$
a_{2}=\frac{7}{2} \frac{p_{0}}{V_{0}}=-1.4 \cdot 10^{5} \mathrm{~Pa} / \mathrm{m}^{3}
$$

Based on the work of Tamás Wiandt (Bonyhád, Petőfi S. Gymnasium, IV. grade)

II. solution. The problem can also be solved by using the following statements:

a) Higher isotherms define higher temperature states.

b) When moving from a higher adiabat to a lower adiabat (during any quasi-static process), the system releases heat.

From this, it is clear that to solve the problem, we need to determine the slope of the tangents at the given points. This can be done most simply by differentiation.