Isochoric Process

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Isochoric Process: Definition, Equations, Examples & Importance

Isochoric Process

An Isochoric process is a type of thermodynamic process in which the volume of a system remains constant. This process is characterized by the absence of any change in volume while other properties such as pressure and temperature may vary. In simpler terms, during an isochoric process, the system's volume doesn't expand or contract. Understanding the isochoric process is crucial for analyzing various thermodynamic systems and is an important concept in both theoretical and practical applications.

1.0Definition of Isochoric Process

It is a thermodynamic process that takes place at constant volume of the system, but pressure and temperature varies for change in state of the system.
Because the volume is constant, the work done by or on the system during an isochoric process is zero.
One common example of an isochoric process is the heating of a gas inside a rigid container. Since the container’s volume does not change, any heat added to the gas results in an increase in pressure or temperature, but not in the volume.

2.0Equation of Isochoric Process

Work Equation

$W = \int d W = \int P d V$

W = 0 ,For an isochoric process dV=0 so work done is equal to zero.

By First Law of Thermodynamics

$Δ Q = Δ U + Δ W$

$(∴ Δ W = 0)$

$Δ Q = Δ U$

$\Rightarrow$ where ΔU is the change in internal energy and Q is the heat added to or removed from the system.

Ideal Gas Relationship

PV = nRT

P ∝ T (∴ V = Constant)

$\frac{P}{T} = Constant$

$\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$

where $P_{1}$ and $T_{1}$ are the initial pressure and temperature, and $P_{2}$ and $T_{2}$ are the final pressure and temperature.

Change in Internal Energy for an Ideal Gas:

For an ideal gas, the change in internal energy ΔU is related to the change in temperature by:

$Δ U = n C_{V} Δ T$ where n is the number of moles, $C_{V}$ is the specific heat capacity at constant volume, and ΔT is the change in temperature.

3.0Derivation of Work Done in Isochoric Process

Work Done is represented by,

dW = P dV

$\int d W = \int_{V_{i}}^{V_{f}}$ P d V

( $∴$ d V=0 for Isochoric Process)

W= 0

4.0Graph of Isochoric Process

5.0Slope of the PV Curve

Slope of PV curve $\frac{d P}{d V}$ =∞

6.0Examples of Isochoric Process

Raising the temperature of a gas within a fixed-volume container

Example: Consider a gas enclosed in a rigid, sealed container. If the gas is heated, the volume of the container cannot change, so the process is isochoric. As the gas heats up, its pressure increases according to the ideal gas behaviour, internal energy ( $Δ U$ ) of the gas increases due to the rise in temperature.

Physical Processes in a Rigid Apparatus:

Example: Any laboratory experiment using a rigid apparatus where the volume of the working substance (e.g. a gas) cannot change, and you’re only varying pressure or temperature. The process is isochoric as long as the volume remains unchanged.

7.0Points to Remember in Isochoric Process

1.Constant Volume: The defining characteristic of an isochoric process is that the volume of the system remains constant throughout the process.

2.Work Done is Zero: Since the volume does not change, the work done W during an isochoric process is zero. This follows from the thermodynamic work equation

dW = P dV

$\int d W = \int_{V_{i}}^{V_{f}}$ P d V

( $∴$ d V=0 for Isochoric Process)

W= 0

3.First Law of Thermodynamics Simplification: The first law of thermodynamics simplifies to ΔU=Q because W=0. Here, ΔU represents the change in internal energy, and Q is the heat added to or removed from the system.

4.Pressure-Temperature Relationship (for Ideal Gases):For an ideal gas undergoing an isochoric process, the relationship between pressure and temperature is given by

$\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ , where $P_{1}$ and $T_{1}$ are the initial pressure and temperature, and $P_{2}$ and $T_{2}$ are the final pressure and temperature.

5.Change in Internal Energy: For an ideal gas, the change in internal energy ΔU can be expressed as $Δ U = n C_{V} Δ T$ .

6.Isochoric Specific Heat Capacity: The specific heat capacity at constant volume $C_{V}$ is used to quantify how much heat is required to change the temperature of the gas maintained at a constant volume. It is related to the change in internal energy.

7.No Volume Work: Because the volume is fixed, the process involves no expansion or compression work. Thus, all the heat added to the system goes directly into changing the internal energy of the system.

8.Graphical Representation: On a PV diagram (Pressure vs. Volume diagram), an isochoric process is represented as a vertical line because the volume remains constant while pressure changes.

9.Applicability in Real Systems: While true isochoric processes are idealizations, in practical scenarios, systems that approximate constant volume include rigid containers or closed vessels where volume is not allowed to change.

10.Thermodynamic Efficiency: Since no work is done during an isochoric process, its efficiency and performance in energy conversion are typically considered in contexts where other factors like pressure and temperature changes are more relevant.

8.0Sample Questions in Isochoric Process

Q-1. What is the heat input needed to raise the temperature of 3 moles of monoatomic gas from 0⁰C to 100⁰C at Isochoric Process?

Solution:

$Q = n C_{V} Δ T$

$Q = 3 \times \frac{3 R}{2} \times (100 - 0) = 450 R$

Q-2 The figure shows a graph of density versus pressure for one mole of an ideal monoatomic gas undergoes a cyclic process. Find the work done in process 3 $\to$ 1 and heat rejected by gas?(M=gas molecular mass)

Find the work done in process 3 $\to$ 1 and heat rejected by gas?

Solution:

Process 3 $\to$ 1, density ( $ρ$ ) is constant so it is an isochoric process hence V is also constant.

$W_{3 \to 1} = 0 (∴ ρ = Constant, so V = Constant$ the process is isochoric)

Heat Rejected by the gas

$Δ U_{3 \to 1} = C_{V} Δ T$

$Δ U_{3 \to 1} = (\frac{3}{2} R) (T_{1} - T_{3})$

$Δ U_{3 \to 1} = (\frac{3}{2} R) (\frac{p _{0} M}{ρ _{0} R} - \frac{2 p _{0} M}{ρ _{0} R})$

$Q_{3 \to 1} = Δ U_{3 \to 1} = \frac{- 3 p _{0} M}{2 ρ _{0}}$

Q-3.A cyclic process ABCD is given for helium gas $C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R$ as shown in fig.

Find work done and internal energy in path $C \to D$

Solution:

Process C $\to$ D is an isochoric process because V= Constant

Work Done
$W = \int$ P d V=0

$W_{C \to D} = 0$

Internal Energy ( $Δ U$ ) for path C $\to$ D

$Δ U_{C \to D} = n C_{V} Δ T = n (\frac{3}{2} R) (T_{D} - T_{C})$

$Δ U_{C \to D} = \frac{3}{2} (n R T_{D} - n R T_{C})$

$Δ U_{C \to D} = \frac{3}{2} (P_{D} V_{D} - P_{C} V_{C})$

$Δ U_{C \to D} = \frac{3}{2} (2 P_{0} V_{0} - 6 P_{0} V_{0})$

$Δ U_{C \to D} = \frac{3}{2} (- 4 P_{0} V_{0}) = - 6 P_{0} V_{0}$

$Δ U_{C \to D} = - 6 P_{0} V_{0}$

Q-4.Given the PV diagram of a thermodynamic cycle for an engine using helium gas, determine the total amount of heat extracted from the heat source during one complete cycle.

Solution:

Heat Extracted from the source $= Δ Q_{D A} + Δ Q_{A B}$

$= n C_{V} Δ T_{D A} + n C_{P} Δ T_{A B}$

$= \frac{3}{2} R (T_{A} - T_{D}) + \frac{5}{2} R (T_{B} - T_{A})$

$= \frac{3}{2} (2 P_{0} V_{0} - P_{0} V_{0}) + \frac{5}{2} (2 P_{0} \times 2 V_{0} - 2 P_{0} V_{0})$

= $\frac{3}{2} P_{0} V_{0} + \frac{5}{2} \times 2 P_{0} V_{0} = \frac{13}{2} P_{0} V_{0}$

Q-5.If the ratio of the specific heat capacities of a gas is given, Calculates the change in internal energy of a gas when its volume increases from V to 2V at constant pressure P.

Solution:

$γ = \frac{C _{P}}{C _{V}} = Sp ec i f i cHe a tR a t i o$

$γ - 1 = \frac{C _{P}}{C _{V}} - 1 = \frac{C _{P} - C _{V}}{C _{V}}$

$C_{V} = \frac{C _{P} - C _{V}}{γ - 1} = \frac{R}{γ - 1} (∴ C_{P} - C_{V} = R)$

$Δ U = n C_{V} Δ T = \frac{n R Δ T}{γ - 1}$

$Δ U = \frac{n P Δ V}{γ - 1} = \frac{n P ( 2 V - V )}{γ - 1} = \frac{n P V}{γ - 1} (∴ P Δ V = n R Δ T)$

For n=1

$Δ U = \frac{P V}{γ - 1}$

Isochoric Process

An Isochoric process is a type of thermodynamic process in which the volume of a system remains constant. This process is characterized by the absence of any change in volume while other properties such as pressure and temperature may vary. In simpler terms, during an isochoric process, the system's volume doesn't expand or contract. Understanding the isochoric process is crucial for analyzing various thermodynamic systems and is an important concept in both theoretical and practical applications.

1.0Definition of Isochoric Process

It is a thermodynamic process that takes place at constant volume of the system, but pressure and temperature varies for change in state of the system.
Because the volume is constant, the work done by or on the system during an isochoric process is zero.
One common example of an isochoric process is the heating of a gas inside a rigid container. Since the container’s volume does not change, any heat added to the gas results in an increase in pressure or temperature, but not in the volume.

2.0Equation of Isochoric Process

Work Equation

$W = \int d W = \int P d V$

W = 0 ,For an isochoric process dV=0 so work done is equal to zero.

By First Law of Thermodynamics

$Δ Q = Δ U + Δ W$

$(∴ Δ W = 0)$

$Δ Q = Δ U$

$\Rightarrow$ where ΔU is the change in internal energy and Q is the heat added to or removed from the system.

Ideal Gas Relationship

PV = nRT

P ∝ T (∴ V = Constant)

$\frac{P}{T} = Constant$

$\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$

where $P_{1}$ and $T_{1}$ are the initial pressure and temperature, and $P_{2}$ and $T_{2}$ are the final pressure and temperature.

Change in Internal Energy for an Ideal Gas:

For an ideal gas, the change in internal energy ΔU is related to the change in temperature by:

$Δ U = n C_{V} Δ T$ where n is the number of moles, $C_{V}$ is the specific heat capacity at constant volume, and ΔT is the change in temperature.

3.0Derivation of Work Done in Isochoric Process

Work Done is represented by,

dW = P dV

$\int d W = \int_{V_{i}}^{V_{f}}$ P d V

( $∴$ d V=0 for Isochoric Process)

W= 0

4.0Graph of Isochoric Process

5.0Slope of the PV Curve

Slope of PV curve $\frac{d P}{d V}$ =∞

6.0Examples of Isochoric Process

Raising the temperature of a gas within a fixed-volume container

Example: Consider a gas enclosed in a rigid, sealed container. If the gas is heated, the volume of the container cannot change, so the process is isochoric. As the gas heats up, its pressure increases according to the ideal gas behaviour, internal energy ( $Δ U$ ) of the gas increases due to the rise in temperature.

Physical Processes in a Rigid Apparatus:

Example: Any laboratory experiment using a rigid apparatus where the volume of the working substance (e.g. a gas) cannot change, and you’re only varying pressure or temperature. The process is isochoric as long as the volume remains unchanged.