Solved Example for First Law of Thermodynamics for a Closed System #1

Question 1 

A horizontal, perfectly insulated piston-cylinder device contains 1.6mol of helium at 800 kPa. Its volume is 0.005m³ and the ambient pressure is 100 kPa.

a) Calculate the work done when the gas is expanded reversibly until the internal pressure is equal to the ambient pressure. Also sketch the process on a P - V diagram.

b) Calculate the work done when the gas is expanded very suddenly until the internal pressure is equal to the ambient pressure.

Answer:

Summary Reversible Processes Involving an Ideal Gas

The table below summarizes the equations studied in the previous blog entry topics on reversible processes involving an ideal gas.

Reversible Processes Involving an Ideal Gas Part 2

Constant Volume (Isometric or isochoric) Process

Changes in internal energy and enthalpy are given by:

For a constant volume process, the work associated with the displacement of system boundaries is zero. Besides, if shaft work is zero or negligible, then W = 0. The amount of transferred heat can be found from the application of the first law of thermodynamics:

Constant Pressure (Isobaric) Process

The changes in internal energy and enthalpy are given by:

Furthermore, the heat transferred and work done can be found using the following:

Constant Temperature (Isothermal) Process

Since temperature remains constant, then:

The work done in a reversible process and heat transferred are found by the following relationships:

Adiabatic Process

An adiabatic process is one in which the amount of heat transferred between the system and its surroundings is zero, i.e., Q = 0

The changes in internal energy and enthalpy are given by:

The work done in a reversible process can be calculated from:

Since temperature changes in an adiabatic process, evaluation of this integral is only possible numerically. Thus, it is more convenient to calculate work by the application of the first law, i.e.,

In some problems, we may know the values of either pressure or volume at the final state. Thus, it is necessary to develop equations between T and P, T and V , and P and V.

We can substitute the ideal gas equation of state below into the above equation to eliminate volumes terms and get a relationship between P and T. The same can also be done to eliminate T and get an relationship between P and V.

Polytropic Process

The relationship between pressure and volume during compression and expansion of gases is sometimes expressed as:

where n is a constant for any given change. Such processes are called polytropic and the equations available for adiabatic processes can also be used for polytropic ones by replacing γ with n.

Reversible Processes Involving an Ideal Gas

An Ideal gas satisfies the following conditions:

• The equation of state is given by; PV = nRT
• Internal Energy is dependent only on temperature.

Enthalpy of an ideal gas is defined as:

This indicates that enthalpy is also only dependent on temperature.

For a single-phase and single-component system, Gibbs phase rule gives the degrees of freedom as two.

Thus, the state of such system is specified by two independent intensive properties. The dependence of internal energy on temperature and volume leads to a convenient relationship to use in calculating internal energy change. The total differential internal energy is given by:

For an ideal gas, internal energy is independent of volume, thus the above equation can be simplified as follows:

The above two equations are valid for an ideal gas regardless of what kind of process is considered.

The dependence of enthalpy on temperature and pressure leads to a convenient relationship to use in calculating enthalpy change. The total differential of enthalpy is given by:

For an ideal gas, enthalpy is independent of pressure, thus the above equation can be simplified as follows:

The above two equations are valid for an ideal gas regardless of what kind of process is considered.

The relationship between the two heat capacities can be found by differentiation of the enthalpy of an ideal gas equation with respect to temperature as:

       

     

  

  

Reversible Processes in a Closed System

 If the changes in kinetic and potential energies are negligible, the first law of thermodynamics can be expressed in differential form as:

dU = dQ + dW

For a reversible process, in the absence of a shaft work the above equation can be simplified as follows:

dU = dQ – PdV

Constant Volume (Isometric or Isochoric) Process

In the case where dV = 0, the above first law of thermodynamics expression simplifies further into:

dU = dQ

Integrating this expression for a reversible isometric process we obtain the following:

Q = ΔU

For a constant volume process, we know that: 

This equation is developed for a closed system undergoing a reversible isometric process. But since U is a state function, the above equation is applicable to all processes where V₁ = V₂ as shown in the following graph:

Constant Pressure (Isobaric) Process

In the case where Pressure is constant, the first law of thermodynamics expression simplifies into:

dU = dQ – d(PV)

or,

dQ = dH

Integrating the above equation for a reversible isobaric equation we obtain;

Q = ΔH

For a constant pressure process we know that;

This equation is developed for a closed system undergoing a reversible isobaric process. However, since H is a state function the ΔH equation becomes applicable to all processes in which P₁ = P₂.

First Law of Thermodynamics for a Closed System

Conservation of Energy

Internal energy (U) is associated with microscopic motions and forces. Since this energy cannot be seen, it is usually separated from the macroscopic i.e., measurable mechanical energy in order to express the total energy of the system as:

E_total = U + Kinetic Energy (E_K) + Potential Energy (E_P)

Let us consider a closed system to which energy in the form of heat and work is supplied from its surroundings:

The first law of thermodynamics states that the total energy of the universe is constant, i.e.,

E_total (universe) = constant    or     Δ E_total (universe) = 0

 

The universe is composed of the system and its surroundings; thus, the above expression can be expressed as follows:

E_total (system) + E_total (surroundings) = 0

The increase in the total energy of the system is given by:

ΔE_total (system) = ΔU + ΔE_K + ΔE_P

Conversely, the decrease in the total energy of the surroundings is given by:

ΔE_total (surroundings) = -Q – W (where Q is heat into or out of the surroundings and W is work done on or by the surroundings)

Combining the above two equations we obtain the first law of thermodynamics for a closed system:

ΔU + ΔE_K + ΔE_P = Q + W

The differential form of this equation is as follows:

dU + dE_K + dE_P = dQ + dW

(Note: Since Q and W are path functions, these quantities in differential form are expressed as δQ and δW in some examples)

If the changes in kinetic and potential energies are negligible the equation can be simplified as:

ΔU = Q + W

The term W includes expansion and non-expansion type of work. Expansion (or contraction) work is related to the change in the volume of system. While non-expansion work includes shaft work (work done on the system by a rotating mechanical device), chemical work, electrical work, etc.