
Concept explainers
(a)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Work done on the system is positive whereas the work done by the system is negative.
(a)

Answer to Problem 155MP
For free expansion process:
Explanation of Solution
Given information:
Initial pressure = 2.00 atm
Initial volume = 2.00 L
Final pressure = 1.00 atm
Final volume = 4.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
Also, since it is given that the gas is undergoing free expansion, thus, there is no external pressure against the gas.
Thus, work is also equal to zero.
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is also equal to zero.
Thus,
Hence, for free expansion process:
(b)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(b)

Answer to Problem 155MP
For expansion done in one step:
Explanation of Solution
Given information:
Initial pressure = 2.00 atm
Initial volume = 2.00 L
Final pressure = 1.00 atm
Final volume = 4.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
Now, work against an external pressure is calculated as:
Put the values,
Thus, work done in one step expansion is
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change in one step expansion is
Hence, for expansion done in one step;
(c)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(c)

Answer to Problem 155MP
For expansion done in two steps;
Explanation of Solution
Given information:
Initial pressure = 2.00 atm
Initial volume = 2.00 L
Final pressure = 1.00 atm
Final volume = 4.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
Now, work against an external pressure is calculated as:
Pressure at the end of the first step is calculated as:
Put the values,
Work done for each step is calculated as:
Now, total work done is calculated as:
Total work done =
=
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change for expansion intwo steps is
Hence, for expansion done in two steps;
(d)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(d)

Answer to Problem 155MP
For expansion done in reversible;
Explanation of Solution
Given information:
Initial pressure = 2.00 atm
Initial volume = 2.00 L
Final pressure = 1.00 atm
Final volume = 4.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
The mathematical expression for work done for a reversible process (expansion) is:
Ideal gas equation is:
Put the values,
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change when expansion is reversible is
Hence, for expansion is reversible;
(e)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(e)

Answer to Problem 155MP
For compression done in one step;
Explanation of Solution
Given information:
Initial pressure = 1.00 atm
Initial volume = 4.00 L
Final pressure = 2.00 atm
Final volume = 2.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
The mathematical expression for work done is:
Put the values,
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change for compression in one step is
Hence, for compression done in one step;
(f)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(f)

Answer to Problem 155MP
For compression done in two steps;
Explanation of Solution
Given information:
Initial pressure = 1.00 atm
Initial volume = 4.00 L
Final pressure = 2.00 atm
Final volume = 2.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
The mathematical expression for work done is:
The work done for each step is calculated as:
Put the values,
Total work done =
=
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change for compression in two steps is
Hence, for compression in two steps;
(g)
Interpretation:
The value of w, q,
Concept Introduction:
The energy which is equal to the sum of kinetic and potential energy of the particles in system is known as internal energy of the system.
The mathematical expression of internal energy is:
Where, E = internal energy, q = heat and w = work done by or on the system
Thermodynamic work done is equal to
Work done on the system is positive whereas the work done by the system is negative.
(g)

Answer to Problem 155MP
For compression is reversible;
Explanation of Solution
Given information:
Initial pressure = 1.00 atm
Initial volume = 4.00 L
Final pressure = 2.00 atm
Final volume = 2.00 L
Since, it is given that gas is an ideal gas.
Both Internal energy and enthalpy depends upon temperature.
Therefore,
In case of isothermal process, the change in internal energy and change in enthalpy of an ideal gas is equal to zero as the temperature is constant.
The mathematical expression for work done for a reversible process (compression) is:
Ideal gas equation is:
Put the values,
Now, change in internal energy is equal to the sum of work done and heat.
Since, change in internal energy is equal to zero.
Since, work done is equal to
Thus,
Therefore, heat change when compression is reversible is
Hence, for compression is reversible;
From above calculations, it is clear that more heat is released for the irreversible compression steps in comparison to irreversible expansion steps.
Since, the system returned to its original state; thus,
Now, for reversible process; during expansion, heat added to the system is equal to the heat lost during compression process. Thus,
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