The branch of thermodynamics studies heat exchange and temperature in gas systems.
- It also examines how heat flows within a system.
- Real-world systems have parameters called "thermodynamic variables," which help describe their state and predict their behavior.
A system is in a thermodynamic state of equilibrium when its macroscopic variables, such as pressure, temperature, volume, mass, and composition, remain constant over time.
Example: A gas stored in a fully insulated container with fixed pressure, volume, temperature, and mass is in equilibrium. Each equilibrium state can be described by these state variables.
In short, thermodynamic variables describe the state of the system at equilibrium. These various state variables are not necessarily independent.
These variables can be divided into two types:
1. Extensive Variables
These variables are the state variables that indicate the size of the system. For example, Volume can be considered an extensive variable because it gives us an idea about the size of the system.
2. Intensive Variables
These variables are the state variables that do not give us any information about the size of the system but indicate different information about the system. Examples of such variables are pressure, temperature, etc.
State Equation
The state equation describes the relationship between the state variables of a thermodynamic system, typically pressure, temperature, and volume. For an ideal gas, it is given by:
PV = RT or PV = Constant
In the case of an isothermal process, where temperature remains constant:
P1V1 = P2V2
These equations define the behavior of ideal gases under different conditions.
Sample Problems
Question 1: In an isothermal thermodynamic process, the initial pressure and volume are 106 N/m²2 and 3 m,3 respectively. Now, the pressure of the container is doubled. Find the volume.
Solution: In the case of an isothermal process,
P1V1 = P2V2
Given:
P1 = 106, P2 = 2 × 106 and V1 = 3
Plugging the values in the equation,
P1V1 = P2V2
⇒ 106 × 3 = 2 × 106 × V2
⇒ 3 = 2 × V2
⇒ 1.5 m3 = V2
Question 2: In an isothermal thermodynamic process, initial pressure and volume are 5 × 106 N/m2 and 6m3 respectively. Now, the pressure of the container is halved. Find the volume.
Solution: In the case of an isothermal process,
P1V1 = P2V2
Given:
P1 = 5 × 106, P2 = 2.5 × 106 and V1 = 6
Plugging the values in the equation,
P1V1 = P2V2
⇒ 5 × 106 × 6= 2.5 × 106 × V2
⇒ 15 x 6 = 2.5 × V2
⇒ 12m3 = V2
Question 3: In an isochoric process, the volume remains constant. Initial pressure and temperature are 5 × 106 N/m2 and 100K, respectively. Now, the pressure of the container is halved. Find the new temperature.
Solution: In the case of an isothermal process,
P1T2 = P2T1
Given:
P1 = 5 × 106, P2 = 2.5 × 106 and T1 = 100 K
Plugging the values in the equation,
P1T2 = P2T1
⇒ 5 × 106 × T2 = 2.5 × 106 × 100
⇒ T2 = 0.5 × 100
⇒ 50 K = T2
Question 4: In an isochoric process, the volume remains constant. Initial pressure and temperature are 106 N/m2 and 250K, respectively. Now, the pressure of the container is increased four times. Find the new temperature.
Solution: In the case of an isothermal process,
P1T2 = P2T1
Given:
P1 = 106, P2 = 4 × 106 and T1 = 250 K
Plugging the values in the equation,
P1T2 = P2T1
⇒ 1 × 106 × T2 = 4 × 106 × 250
⇒ T2 = 4 × 250
⇒ 1000K = T2
Question 5: In a thermodynamic process, the pressure remains constant. Initial volume and temperature are 5 m³3 and 250K, respectively. Now, the volume of the container has increased by two times. Find the new temperature.
Solution: In the case of an isothermal process,
V1T2 = V2T1
Given:
V1 = 5 m3, V2 = 10 m3 and T1 = 250 K
Plugging the values in the equation,
V1T2 = V2T1
⇒ 5 × T2 = 10 × 250
⇒ T2 = 500
⇒ 500K = T2
Unsolved Problems
Question 1: In an isothermal process, a gas has an initial volume of 4 m³ and a pressure of 2 × 105 N/m². If the gas is compressed to 2 m³, find the final pressure.
Question 2: In an isochoric process, the pressure of a gas is 3 × 10²5 N/m² at 300 K. If the pressure is increased to 4.5×105 N/m², find the new temperature.
Question 3: A gas in an isobaric process has an initial volume of 2 m³ and a temperature of 200 K. If the volume is increased to 5 m³, calculate the new temperature.
Question 4: In an isothermal process, the volume of a gas decreases from 6 m³ to 3 m³ while the initial pressure is 1.5 × 105 N/m². Find the final pressure.
Question 5: A gas is in an isochoric process at 400 K and a pressure of 5 × 105 N/m². If the temperature is raised to 600 K, calculate the new pressure.