Name: Sign Conventions and Work Done in Thermodynamics
Uploaded: 2019-07-19
Duration: 6 min 11 s
Description: In the field of thermodynamics, either heat is given to a system or by the system. Similarly, work may be done by the system or on the system. Let's learn more!

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 monatomic 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 (Fig.).

Determine the heat capacity $C$ of the system, neglecting the heat capacities of the vessel, piston, and spring.

Accepted Answer

Let $T_{1}$ be the initial temperature of the gas under the piston, and $T_{2}$ the gas temperature after the amount of heat $Δ Q$ has been supplied to the system. Since there is no friction and the vessel is thermally insulated, the entire amount of heat $Δ Q$ is spent on the change in the internal energy of the system, $\Rightarrow Δ Q = Δ U$

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 monatomic gas increases as a result of heating from $T_{1}$ to $T_{2}$ by, $Δ U_{1} = \frac{3}{2} R (T_{2} - T_{1}) . . . . (1)$

The potential energy of the compressed spring changes by, $Δ U_{2} = \frac{k}{2} (x_{2}^{2} - x_{1}^{2}) . . . . . (2)$

where $k$ is 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 temperatures $T_{1}$ and $T_{2}$ respectively. Let us 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

$p = \frac{F}{S} = \frac{k x}{S}, x = \frac{p S}{k} . . . . (3)$ ,

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 $p V = R T$ . For the deformation $x$ of the spring, the volume of the gas under the piston is $V = x S$ and the pressure $p = \frac{R T}{x S}$ . Substituting this expression for $p$ into Eq. $(3)$ , we obtain

$x^{2} = \frac{R T}{k} . . . . (4)$

Thus, the change in the potential energy of the compressed spring as a result of heating of the system is $Δ U_{2} = \frac{R}{2} (T_{2} - T_{1}) .$

The total change in the internal energy of the system as a result of heating from $T_{1}$ to $T_{2}$ is
$Δ U = Δ U_{1} + Δ U_{2} = 2 R (T_{2} - T_{1})$ , and the heat capacity of the system is

$C = \frac{Δ Q}{Δ T} = \frac{Δ U}{T_{2} - T_{1}} = 2 R .$

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