Question

A thermally insulated vessel is divided into two parts by a heat-insulating piston which can move in the vessel without friction. The left part of the vessel contains one mole of an ideal monoatomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel. Heat capacity $C$ of the system $=XR$. Find $x$? (neglecting the heat capacities of the vessel, piston and spring).

Answer 1

The system with heat capacity $C=XR$

Find $x=?$

Solution:-

Let, $T_1$ be the initial temperature of the gas under the piston, and $T_2$ the gas temperature after the amount of heat $\Delta Q$ has been supplied to the system. Since there is no friction and the vessel is thermally insulated, the entire amount of heat $\Delta Q$ is spent. On the change $\Delta W$ in the internal energy of the system

$\Delta Q=\Delta W$

The change in the internal energy of the system is the sum of the changes in the internal energy of the gas and in the potential energy of the compressed spring (Since we neglect the heat capacity of the vessel, piston and spring). The internal energy of a mole of an ideal monoatomic gas increases as a result of heating from $T_1$ $T-2$ by

$\Delta W_1=\frac{3}{2}R(T_2-T_1)\to \left(1\right)$

The potential energy of the compressed spring changes by

$\Delta W_2=\frac{K}{2}(x_2^2-x_1^2)\to \left(2\right)$

Where, $K$ us the rigidity of the spring, and $x_1$ and $x_2$ are the values of the absolute displacement (deformation) of the left end of the spring at temperature $T-1$ and $T_2$ respectively.

Letus find the relation between the parameters of the gas under the piston and the deformation of the spring.

The equilibrium condition for the piston implies that

$P=\frac{F}{S}=\frac{Kx}{S},x=\frac{PS}{K}\to \left(3\right)$

Where P is the gas pressure and S is the area of the piston. According to the equation of state for an ideal gas , for one mole we have $PV=RT$ for the deformation $x$ of the spring, the volume of the gas under the piston is $V=KS$ and pressure $P=\frac{RT}{XS}$

Substituting this expression for P into equation $(3$ we obtain

$x^2=\frac{RT}{K}\to 4$

Thus the change in the potential energy of the compressed spring as result of heating of the system is$\Delta W_2=\frac{R}{2}(T_2-T_1)$

The total charge in the internal energy of the system as a result of heating from $T_1$ $T_2$ is

$\Delta W=\Delta W_1+\Delta W_2=2R(T_2-T_1)$

And the heat capacity of the system is

$C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{T_2-T_1}=2R$

Now,

$XR=2RX=2$