Question

\(12-104 \quad\) For a gas whose equation of state is \(P(v-b)=R T\) the specified heat difference \(c_{p}-c_{v}\) is equal to \((a) R\) (b) \(R-b\) \((c) R+b\) \((d) 0\) \((e) R(1+v / b)\)

Short answer

In this problem, we are given the equation of state for a gas: \(P(v-b)=RT\). We applied the first law of thermodynamics for ideal gas and found expressions for specific heat at constant volume (\(c_v\)) and constant pressure (\(c_p\)). After calculating the difference between \(c_p\) and \(c_v\), we found that the difference is equal to the universal gas constant (\(R\)). Therefore, the answer is \(c_p - c_v = R\).

Step-by-step solution

First law of thermodynamics for ideal gas

We will use the first law of thermodynamics for an ideal gas, which states that \(dU = nC_v dT = nPdV\) in case of constant volume and \(dH = nC_p dT = nVdP\) in case of constant pressure. Here, U represents the internal energy, H represents the enthalpy and n is the number of moles.

Using differential equations

We can rewrite the first law of thermodynamics for these 2 situations: For constant volume, \((1) \quad C_v = \frac{PdV}{dT}\) For constant pressure \((2) \quad C_p = \frac{VdP}{dT}\) Now, we have to find expressions for \(C_v\) and \(C_p\) in terms of the given equation of state.

Differentiate the equation of state

The given equation of state, \(P(v-b) = RT\), can be differentiated with respect to T keeping the volume constant: \(d(Pv-Pb)=RdT\) Rewriting this equation as: \(d(Pv)=Pd(v-b)+RdT\) Dividing both sides by \(dT\), we get \( \frac{Pd(v)}{dT} = P\frac{d(v-b)}{dT}+R\) Now, we can substitute this back into equations (1) and (2) and solve for \(C_p\) and \(C_v\).

Find expressions for \(C_v\) and \(C_p\)

Substituting the new expression into equations (1) and (2), we get: For constant volume \(C_v = P\frac{d(v-b)}{dT}\), \((3)\) For constant pressure, \(C_p = \frac{Pd(v-b)+R}{dT}\), \((4)\) We can rearrange equation (4) to obtain the expression for \(C_p\): \(C_p -C_v = \frac{R}{dT}\), \((5)\) Now we have an expression for the difference between \(C_p\) and \(C_v\).

Calculate \(c_p - c_v\)

From equation (5), we find the difference between \(c_p\) and \(c_v\): \(c_p - c_v = \frac{R}{dT}\) Comparing with the given options, the closest match is (a) \(R\). Therefore, \(c_p - c_v = R\).

Key concepts

Thermodynamics Equation of State

The equation of state in thermodynamics is a mathematical formula that relates the pressure, volume, and temperature of a gas. For most ideal gases, the equation of state is expressed as PV = nRT, where P stands for pressure, V for volume, T for temperature, n for the number of moles, and R for the ideal gas constant.
In more complex scenarios, real gases exhibit behavior that deviates from the ideal gas law. This is often accounted for by adjusting the equation of state to include additional parameters that reflect intermolecular forces and the volume occupied by the molecules themselves. An example is the Van der Waals equation and the exercise's equation of state, P(v - b) = RT, where b is a constant representing the volume occupied by the gas molecules.
The equation of state is crucial as it allows for the derivation of other important thermodynamic properties and indices, like specific heats, through differentiation and manipulation.

Specific Heats of Gases

Specific heats of gases refer to the amount of heat required to raise the temperature of a gas by one degree at constant volume (c_v) or at constant pressure (c_p). The specific heats are critical in understanding how gases absorb and transfer heat during thermodynamic processes.
For an ideal gas, the difference between the two specific heats, c_p and c_v, is equal to the gas constant, R. This relationship holds true because of the work done by or on the gas during expansion or compression at constant pressure, which does not occur at constant volume. This concept is applied in the textbook solution, leading to the result that c_p - c_v equals R. The context of the gas's equation of state can complicate this, but the core idea remains a cornerstone of thermodynamics.

Thermodynamic Processes

Thermodynamic processes describe the pathways by which a system changes from one state to another. Common processes include isobaric (constant pressure), isochoric (constant volume), isothermal (constant temperature), and adiabatic (no heat transfer).
During these processes, variables such as pressure, volume, and temperature change in ways that can be described by the first law of thermodynamics. This law essentially states that the energy within a closed system is conserved. When applied to thermodynamic processes, it helps us determine the work done by or on the system and the heat exchange involved. By understanding these processes and their implications on energy changes, students can solve complex problems in thermodynamics, like the one described in the textbook exercise, using equations derived from the first law and the applicable equation of state.