For an ideal gas, internal energy depends only on temperature.
\[ \Delta U = n C_V \Delta T \]
Given:
\[ C_P = 7 \text{ cal/mol °C} \]
\[ R = 2 \text{ cal/mol °C} \]
Using relation:
\[ C_P - C_V = R \]
\[ 7 - C_V = 2 \]
\[ C_V = 5 \text{ cal/mol °C} \]
\[ n = 10 \]
\[ \Delta T = 10°C \]
\[ \Delta U = 10 \times 5 \times 10 = 500 \text{ cal} \]
Correct Answer: 500 cal
Thermodynamics is one of the most conceptually rich and highly scoring chapters in JEE Main and JEE Advanced. Understanding internal energy, heat capacities and the first law is absolutely essential.
Internal energy of an ideal gas depends only on temperature.
This is because ideal gas assumes no intermolecular potential energy.
All internal energy is purely kinetic.
Thus internal energy change depends only on change in temperature.
The general expression is:
\[ U = \frac{f}{2} nRT \]
Here f is degrees of freedom.
For monoatomic gas, f = 3.
For diatomic gas at room temperature, f = 5.
Oxygen is diatomic.
Thus f = 5.
Hence:
\[ C_V = \frac{f}{2}R \]
\[ C_V = \frac{5}{2}R \]
Given R = 2 cal/mol°C.
\[ C_V = 5 \text{ cal/mol°C} \]
Heat capacity at constant pressure:
\[ C_P = C_V + R \]
This is an extremely important formula.
Students must memorize:
\[ C_P - C_V = R \]
This relation is universal for ideal gases.
At constant volume process:
\[ W = 0 \]
Because work done:
\[ W = \int P dV \]
If volume constant, dV = 0.
So W = 0.
From first law:
\[ Q = \Delta U + W \]
Thus:
\[ Q = \Delta U \]
Hence heat supplied equals internal energy change.
Temperature difference must always be calculated carefully.
30°C to 40°C means ΔT = 10°C.
For ideal gas, Celsius and Kelvin difference are same.
So conversion not needed here.
Common mistakes:
Using Cp instead of Cv.
Forgetting Cp - Cv relation.
Wrong ΔT calculation.
Multiplication mistakes.
Confusing heat with internal energy.
In JEE Main, such direct formula questions are common.
But sometimes twist is given in process type.
For isothermal process:
\[ \Delta U = 0 \]
For adiabatic process:
\[ Q = 0 \]
For isobaric process:
\[ Q = n C_P \Delta T \]
For isochoric process:
\[ Q = n C_V \Delta T \]
Students should clearly distinguish between these.
Degrees of freedom concept is also important.
Each translational motion contributes 1 degree.
Each rotational motion contributes 1 degree.
Vibrational contributes 2 degrees.
At room temperature, vibration is inactive.
So diatomic gas has 5 degrees.
Equipartition theorem states each degree contributes:
\[ \frac{1}{2} kT \]
per molecule.
Thus internal energy is proportional to T.
This explains why internal energy change depends only on temperature.
Thermodynamics questions from Cp, Cv appear almost every year.
They are easy scoring if concepts are clear.
Revision tip:
Always remember Cp = Cv + R.
Remember monoatomic values: Cv = 3/2 R.
Diatomic: Cv = 5/2 R.
Polyatomic: Cv = 3R (approx).
Practice numerical substitution carefully.
Never skip units.
Be careful with calories and Joules conversion if needed.
This question directly tests conceptual clarity.
Mastering these basics builds strong foundation.
1. What is internal energy?
It is total microscopic kinetic energy of molecules.
2. Why independent of volume?
Because ideal gas has no potential energy.
3. What is Cp - Cv?
Equal to R.
4. Oxygen degrees of freedom?
5 at room temperature.
5. Why work zero?
Volume constant.
6. ΔT in Kelvin or Celsius?
Either, difference same.
7. Chapter name?
Thermodynamics.
8. Common trap?
Using Cp.
9. Appears in JEE Main?
Yes frequently.
10. Important formula?
ΔU = nCvΔT.
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