Consider a piston sliding without friction in a cylinder of surface A, which is attached to a spring of elastic constant k (Fig. 2.9). When the cylinder is empty, the piston is at position x0. The cylinder is filled with a gas that satisfies the law pV = NRT. The internal energy of the gas is given

Question

Consider α piston sliding without friction in a cylinder of surface A, which is attached to a spring of elastic constant k (Fig. 2.9). When the cylinder is empty, the piston is at position [latex]x_0[/latex]. The cylinder is filled with a gas that satisfies the law pV = NRT. The internal energy of the gas is given by U = cNRT where c > 0 and R > 0 are constants. After filling the cylinder with gas, the piston is at equilibrium at the initial position [latex]x_i[/latex]. Then, the cylinder heats up and reaches the final equilibrium position at [latex]x_f[/latex]. The process is assumed to be reversible and the system is in a vacuum chamber, i.e. the pressure vanishes outside the system. The mass of the piston is not taken into consideration here.
1. Determine the volume [latex]V_α[/latex], pressure [latex]p_α[/latex] and temperature [latex]T_α[/latex] of the gas in any equilibrium state a in terms of the parameters k, A, [latex]x_0[/latex] and [latex]x_α[/latex].
2. Show that the derivative of the pressure p with respect to the volume V is given by,

[latex]\frac{dp}{dV}=\frac{k}{A^2}[/latex].

3. Determine the work [latex]−W_{if}[/latex] performed by the gas on the spring when the piston moves from [latex]x_i[/latex] to [latex]x_f[/latex] in terms of the parameters [latex]k, x_i[/latex] and [latex]x_f[/latex].
4. Determine the internal energy variation [latex]ΔU_{if}[/latex] of the gas when the piston moves from [latex]x_i to x_f[/latex] in terms of the parameters[latex] k, c, x_0, x_i[/latex] and [latex]x_f[/latex].

Accepted Answer

1. In the equilibrium state α, where α ∈ {i, f}, the volume of the ideal gas is given by,

[latex]V_α=Ax_α[/latex].

and the pressure of the ideal gas is given by

[latex]pα=-\frac{F_α}{A} =\frac{k}{A} (x_α-x_0)[/latex].

where [latex]F_α[/latex] is the projection on the x-axis of the elastic force acting on the spring. Given that the process is reversible, relation (2.35)

[latex]F^{cont}=p^{ext}A[/latex]

[latex]\dot{V}=-A.ν[/latex].

implies that the pressure of the ideal gas is equal to the pressure exerted by the spring. The temperature of the ideal gas is given by,

[latex]T_α=\frac{p_αVα}{NR}=\frac{k}{NR} (x_α-x_0)x_α[/latex].

2. The equations for the volume and the pressure imply,

[latex]dp=-\frac{1}{A}dF=\frac{k}{A}dx=\frac{k}{A^2}dV [/latex].

Thus,

[latex]\frac{dp}{dV}=\frac{k}{A^2}=const[/latex].

3. Hence, the pressure p is expressed in terms of the volume V as,

[latex]p=\frac{k}{A^2}(V-V_0)[/latex].

The work −[latex]W_{if}[/latex] performed by the ideal gas on the spring is given in terms of [latex]V_i[/latex] and [latex]V_f[/latex] by,

[latex]-W_{if}=\int_{V_i}^{V_f}{pdV} =\frac{k}{A^2} \int_{V_i}^{V_f}{(V-V_0)dV}=\frac{k}{2A^2} (V^{2}_{f}-V^{2}_{i} )-\frac{k}{A^2} V_0(V_{f}-V_{i} )[/latex].

This result can be recast in terms of x as,

[latex]-W_{if}=\frac{k}{2} (x^{2}_{f}-x^{2}_{i} )-kx_0(x_f-x_i)=\frac{k}{2}\Biggl((x_f-x_0)^2-(x_i-x_0)^2\Biggr)[/latex].

The work performed by the ideal gas on the spring is equal to the variation of the elastic energy of the spring during the compression. In other words, the work of the spring is entirely used to compress the gas. This is a consequence of the fact that the thermal expansion of the gas is a reversible process.

4. The internal energy variation [latex]ΔU_{if}[/latex] is given by,

[latex]ΔU_{if}=cNR(T_f-T_i)=ck \Biggl((x_f-x_0)x_f-(x_i-x_0)x_i\Biggr)[/latex].

Question 2.11.5: Consider a piston sliding without friction in a cylinder of ...

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