Sigma Percentile
JEE Main 2005
LEVELBoard

Animated Solution for Physics - Thermodynamics: A system goes from to via two processes I and II as shown in figure. If and are the changes in internal energies in the processes I and II respectively, then

Select Answer:

Visualized Solution

Visualizing the Processes

  • The system transitions from an initial state to a final state .
  • Two different paths, I and II, are taken to achieve this transition.

Internal Energy as a State Function

  • Internal energy () is a state function.
  • It depends only on the initial and final states of the system, not on the path taken.

Comparing the Two Processes

  • For Process I:
  • For Process II:
  • Therefore,

Final Conclusion

  • The change in internal energy is independent of the path.

The Sigma Insight: First Law of Thermodynamics

Solution Diagram
Imagine you are standing at the base of a mountain (State ) and you want to reach the summit (State ). You have two choices: you can take the steep, direct trail (Process I), or you can take the long, winding scenic route (Process II).
If I ask you, "How much distance did you walk?" your answer will heavily depend on the path you chose. The winding route will rack up far more miles than the direct trail. Distance, in this analogy, is a path function.
But what if I ask you, "How much did your altitude change?" It doesn't matter if you took the steep trail, the winding route, or even if you took a helicopter. Your change in altitude is simply the altitude at the summit minus the altitude at the base. Altitude is a state function.

The Thermodynamic Equivalent

In thermodynamics, we deal with similar concepts. When a gas expands or compresses from an initial state to a final state , it can do so via infinite possible paths on a diagram.
The work done by the gas () is the area under the curve. Just like the distance walked on the mountain, the area under the curve is different for Process I and Process II. Therefore, work is a path function.
Similarly, the heat supplied to the system () also depends on the path taken.

The Magic of Internal Energy

However, the internal energy () of an ideal gas depends solely on its temperature, which is uniquely defined by its pressure and volume at any given point ().
Because internal energy only cares about where the system is, and not how it got there, it is a state function.
For Process I, the change in internal energy is:
For Process II, the change in internal energy is:
Since both processes start at the exact same state and end at the exact same state , the change in internal energy must be identical.
Therefore, we can confidently conclude:
This simple yet profound realization is the cornerstone of the First Law of Thermodynamics!

Similar Questions

JEE Main 2019
LEVELJEE Main

Following figure shows two processes A and B for a gas. If and are the amount of heat absorbed by the system in two cases, and and are changes in internal energies respectively, then

(A)
(B)
(C)
(D)
JEE Advanced 2014
LEVELJEE Advanced

A thermodynamic system is taken from an initial state with internal energy to the final state along two different paths and , as schematically shown in the figure. The work done by the system along the paths , and are , and respectively. The heat supplied to the system along the path , and are , and respectively. If the internal energy of the system in the state is and , the ratio is

JEE Main 2003
LEVELJEE Main

The internal energy change when a system goes from state to is . If the system goes from to by a reversible path and returns to state by an irreversible path, what would be the net change in internal energy?

(A)
(B)
(C)
(D)
zero
JEE Advanced 2018
LEVELJEE Advanced

A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

* Multiple Correct Options
(A)
and
(B)
and
(C)
and
(D)
and
JEE Main 2019
LEVELJEE Main

A gas can be taken from to via two different processes and . When path is used 60 J of heat flows into the system and 30 J of work is done by the system. If path is used work done by the system is 10 J the heat flow into the system in path is

(A)
80 J
(B)
40 J
(C)
100 J
(D)
20 J
LEVELJEE Main

For an ideal gas

* Multiple Correct Options
(A)
the change in internal energy in a constant pressure process from temperature to is equal to , where is the molar heat capacity at constant volume and the number of moles of the gas
(B)
the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process
(C)
the internal energy does not change in an isothermal process
(D)
no heat is added or removed in an adiabatic process
LEVELJEE Main

Consider the reaction, carried out at constant temperature and pressure. If and are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

(A)
(B)
(C)
(D)
JEE Main 2014
LEVELJEE Main

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K, respectively. Choose the correct statement.

(A)
The change in internal energy in whole cyclic process is
(B)
The change in internal energy in the process CA is
(C)
The change in internal energy in the process AB is
(D)
The change in internal energy in the process BC is
JEE Main 2017
LEVELJEE Main

is equal to

(A)
isochoric work
(B)
isobaric work
(C)
adiabatic work
(D)
isothermal work
JEE Main 2019
LEVELJEE Main

A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is . The gas absorbs of heat along the path ab and along the path bc. The work done by the gas along the path abc is

(A)
(B)
(C)
(D)