3. (6 points) In a horizontal stationary cylindrical vessel closed by a piston of mass $M$, a gas is contained. The gas is heated, and during this process, the piston moves from rest with constant acceleration. The amount of heat supplied to the gas over the time interval $\tau$ is $Q$. Determine the acceleration of the piston. The internal energy of one mole of the gas is $U=c T$. Neglect the heat capacity of the vessel and the piston. The external side of the piston is in a vacuum.
Solution. Since the piston moves with uniform acceleration, the process is isobaric. According to the first law of thermodynamics,

$$
Q=p \Delta V+v c \Delta T
$$

Here $p$ - the pressure of the gas, $V$ - its volume, $v$ - the amount of substance of the gas.

According to the Clapeyron-Mendeleev equation,

$$
p \Delta V=v R \Delta T
$$

According to the theorem of kinetic energy change,

$$
p \Delta V=\frac{M(a \tau)^{2}}{2}
$$

Thus,

$$
v \Delta T=\frac{M(a \tau)^{2}}{2 R}
$$

In the final form, we get

$$
a=\sqrt{\frac{2 Q}{M \tau^{2}\left(1+\frac{c}{R}\right)}}
$$

Answer: $a=\sqrt{\frac{2 Q}{M \tau^{2}\left(1+\frac{c}{R}\right)}}$.

## Grading Criteria

| Task | Points |
| :--- | :---: |
| The participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| The expression for the first law of thermodynamics is written | $\mathbf{1}$ |
| The Clapeyron-Mendeleev equation is written | $\mathbf{1}$ |
| The theorem of kinetic energy change is written | $\mathbf{2}$ |
| Necessary transformations are performed, the final answer is obtained | $\mathbf{2}$ |
| Total points | $\mathbf{6}$ |