6.1. A vertical cylindrical vessel with a cross-sectional area of $S=10 \mathrm{~cm}^{2}$, containing one mole of a monatomic ideal gas, receives an amount of heat $Q=500$ J per second. The vessel is closed at the top by a heavy piston weighing $P=100 \mathrm{N}$. At what speed (in meters per second) does this piston rise if the atmospheric pressure is $P_{0}=10^{5}$ Pa?
Answer: 1. Solution. According to the first law of thermodynamics, we can write $Q \Delta t=\Delta A+\Delta U$ (2), where $\Delta t$ - time interval, $\Delta A$ - work done by the gas, $\Delta U-$ change in internal energy. Since the process is isobaric, the following relationships can be written for work and internal energy:

$$
\Delta A=\left(p_{0}+\frac{P}{S}\right) \Delta V, \Delta U=\frac{3}{2} v R \Delta T=\frac{3}{2}\left(p_{0}+\frac{P}{S}\right) \Delta V
$$

where $v-$ is the amount of substance. Substituting (3) into (2), we get $Q \Delta t=\frac{5}{2}\left(p_{0}+\frac{P}{S}\right) S \Delta h$. From this, for the speed of the piston movement, we get $\frac{\Delta h}{\Delta t}=\frac{Q}{2.5\left(p_{0} S+P\right)}$