Answer to Question #88249 in Molecular Physics | Thermodynamics for Soumya Kanti Jana

Question

34. A horizontal cylindrical tube is divided into four parts with the help of
two fixed partitions A and B and a movable piston C. The tube and the
piston C are made from heat insulating materials whereas the fixed
partitions A and B are made from a conducting material. Volume
itial volume between
the fixed partition B and piston C is V2. These volumes are filled with
an ideal mono-atomic gas at pressure po and temperature To. Outside
the piston C, pressure is also po. Quantity AQ of heat is slowly
transferred through the partition A to the gas trapped between the
partitions A and B. Find temperatures of the gases on both sides of thebetween the fixed partitions A and B is Vi arpartition B and heat transferred through it.

Answer 1

Draw image

Using first law of thermodynamics for volume between AB

$$Q_A=\Delta U_1+Q_B$$

Using first law of thermodynamics for volume between BC

$$Q_B=\Delta U_2+A_2$$

The increment of internal energy for ideal gas between AB

$$\Delta U_1=\frac{3}{2} \nu_1 R \Delta T$$

The increment of internal energy for ideal gas between BC

$$\Delta U_2=\frac{3}{2} \nu_2 R \Delta T$$

Ideal gas between BC produce work

$$A_2=p_0 \Delta V$$

but, from Mendeleev-Clapeyron equation

$$p_0 \Delta V=\nu_2 R \Delta T$$ $$p_0 V_1=\nu_1 R \Delta T$$ $$p_0 V_2=\nu_2 R \Delta T$$

So, we have equation

$$Q_A=\frac{3}{2} \nu_1 R \Delta T+\frac{3}{2} \nu_2 R \Delta T +\nu_2 R \Delta T$$ $$\Delta T= T-T_0$$ $$T=[\frac{Q_A}{p_0(\frac{3}{2}V_1+\frac{5}{2}V_2)}+1]T_0$$

For heat trough partition B

$$Q_B=\Delta U_2+A_2=\frac{5}{2} \frac{p_0 V_2}{RT_0} R \Delta T$$ $$Q_B=\frac{V_2}{\frac{3}{5}V_1+V_2} Q_A$$

Answer

Temperature gases is equal and most be

$$T=[\frac{Q_A}{p_0(\frac{3}{2}V_1+\frac{5}{2}V_2)}+1]T_0$$

Heat trough partition B

$$Q_B=\frac{V_2}{\frac{3}{5}V_1+V_2} Q_A$$