Quick Summary
- Concept Tested: Ratio of differential changes $dU$, $dQ$, $dW$ for an ideal gas.
- Chapter – Subtopic: Kinetic Theory of Gases – Molar Heat Capacities.
- Difficulty: ★★★☆☆ (moderate)
- Estimated Time: 2 minutes
- Key Formula: $$dU=nC_VdT,\quad dQ=nC_pdT,\quad dW=nRdT$$
- Answer: (B) 5 : 7 : 2
- One‑line Reason: With $C_V=\frac{5}{2}R$ and $C_p=\frac{7}{2}R$, the ratio simplifies to $5:7:2$.
The Question
A diatomic ideal gas has molar heat capacities $C_V=\frac{5}{2}R$ and $C_p=\frac{7}{2}R$. The gas is heated at constant pressure. Find the ratio of the differential changes in internal energy ($dU$), heat added ($dQ$), and work done ($dW$):
$$dU : dQ : dW = \; ?$$
(A) 5 : 7 : 3
(B) 5 : 7 : 2
(C) 3 : 7 : 2
(D) 3 : 5 : 2
Quick Answer
The correct choice is (B) 5 : 7 : 2. Using $dU=nC_VdT$, $dQ=nC_pdT$ and $dW=nRdT$ with the given heat capacities leads directly to the ratio $5:7:2$.
Why Other Options Are Incorrect
Option (A): 5 : 7 : 3
The work term is taken as $3R$ instead of the correct $2R$. For a constant‑pressure process, $dW=nRdT$, not $3nRdT$; thus the third entry is overstated.
Option (C): 3 : 7 : 2
The internal‑energy term uses $3R$ while the correct value is $\frac{5}{2}R$ (i.e., $5R$ after clearing denominators). This misrepresents $C_V$ for a diatomic gas.
Option (D): 3 : 5 : 2
Both $dU$ and $dQ$ are incorrectly scaled: $dU$ should involve $5R$, and $dQ$ should involve $7R$, not $3R$ and $5R$ respectively.
Video Solution
Video Solution Coming Soon
Understanding the Concept
The first law of thermodynamics links heat, work, and internal energy: $dQ = dU + dW$. For an ideal gas, internal energy depends only on temperature, giving $dU=nC_VdT$. At constant pressure, the heat supplied equals $nC_pdT$, while the work done is $p\,dV=nRdT$ obtained from the differentiated ideal‑gas equation $pV=nRT$.
Detailed Step-by-Step Solution
Step 1: Write the expressions for each differential quantity
For an ideal gas:
$$dU = nC_V\,dT$$
$$dQ = nC_p\,dT$$
$$dW = nR\,dT$$
These follow from the definitions of molar heat capacities and the ideal‑gas law.
Step 2: Insert the given numerical values for $C_V$ and $C_p$
Given $C_V=\frac{5}{2}R$ and $C_p=\frac{7}{2}R$, substitute:
$$dU = n\left(\frac{5}{2}R\right)dT,\qquad dQ = n\left(\frac{7}{2}R\right)dT,\qquad dW = nR\,dT$$
Step 3: Form the ratio $dU : dQ : dW$
Factor out the common term $n\,dT$:
$$dU : dQ : dW = \frac{5}{2}R : \frac{7}{2}R : R$$
Step 4: Eliminate the fractional coefficients
Multiply each term by 2 to clear denominators:
$$dU : dQ : dW = 5R : 7R : 2R$$
Step 5: Cancel the common factor $R$
Dividing each term by $R$ yields the final simple ratio:
$$dU : dQ : dW = 5 : 7 : 2$$
Final Answer
✅ Ratio $dU : dQ : dW = 5 : 7 : 2$ (Option B)
Essential Formulas for This Topic
$$dU = nC_V dT$$
$$dQ = nC_p dT$$
$$dW = p\,dV = nR dT$$
$$C_p = C_V + R$$
$$pV = nRT$$
Note: For a diatomic ideal gas, $C_V = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$.
Common Mistakes to Avoid
Mistake 1: Swapping $C_V$ and $C_p$
Students often insert $C_p$ for $dU$ and $C_V$ for $dQ$. Remember $dU$ always uses $C_V$ because internal energy is independent of the process.
Mistake 2: Using $p\,dV = nC_p dT$ for work
Work at constant pressure should be expressed via the ideal‑gas law, giving $dW = nR dT$, not $nC_p dT$.
Mistake 3: Forgetting to cancel the common factor $n\,dT$
Leaving $n$ and $dT$ in the ratio leads to unnecessary complexity; they cancel out because every term contains the same factor.
Key Concept Summary
- First law: $dQ = dU + dW$ must hold for any process.
- For ideal gases, $dU$ depends only on temperature: $dU=nC_VdT$.
- At constant pressure, heat added is $dQ=nC_pdT$.
- Work done in a constant‑pressure expansion is $dW=nRdT$ derived from $pV=nRT$.
- Relation between heat capacities: $C_p = C_V + R$.
Golden Rule: Always express $dU$, $dQ$, and $dW$ in terms of $dT$ before forming ratios; common factors cancel automatically.
Frequently Asked Questions
Q: Why does $dU$ depend only on $C_V$ and not on the process?
A: For an ideal gas, internal energy is a function of temperature alone. $C_V$ quantifies the change in internal energy per mole per degree, irrespective of how the temperature change occurs.
Q: Can we use $C_p$ for work calculation?
A: No. Work at constant pressure is derived from $p\,dV$, which becomes $nR dT$ after using the ideal‑gas law. $C_p$ relates heat to temperature, not work.
Q: What if the gas were polyatomic with more degrees of freedom?
A: The values of $C_V$ and $C_p$ would change (e.g., $C_V=\frac{f}{2}R$ where $f$ is the number of degrees of freedom), but the procedure of forming the ratio remains identical.
Q: Does the ratio change if the process is at constant volume?
A: Yes. At constant volume, $dW=0$, so the ratio becomes $dU : dQ : dW = C_V : C_V : 0$, which simplifies to $1 : 1 : 0$.
Prerequisites to Solve This Question
- Understanding of the first law of thermodynamics.
- Familiarity with molar heat capacities $C_V$ and $C_p$ for ideal gases.
- Ability to manipulate the ideal‑gas equation $pV=nRT$.
- Skill in canceling common factors when forming ratios.
After Solving This, You Can:
- ✔ Derive differential energy relations for any ideal‑gas process.
- ✔ Quickly identify the correct heat capacity to use for $dU$, $dQ$, or $dW$.
- ✔ Apply the first law to verify consistency of thermodynamic calculations.
Study Tips for This Topic
- Memorize the relation $C_p = C_V + R$; it saves time in many JEE problems.
- Practice converting work expressions using the ideal‑gas law rather than memorizing $p\,dV$ directly.
- When given ratios, always write each term with its explicit $R$ factor before simplifying.
- Work on numerical variations (different gases, constant volume vs. constant pressure) to reinforce the concepts.
Difficulty Rating & Exam Frequency
Difficulty: ★★★☆☆ (moderate)
JEE Main Frequency: Common – appears in thermodynamics sections.
JEE Advanced Frequency: Occasional – often combined with other concepts like degrees of freedom.
Importance: High – foundational for all later thermodynamic problems.
Related Questions from Kinetic Theory Of Gases
Written by Minal Kumari
Senior Physics Educator at Padho Likho JEE. Specializes in breaking down complex mechanics and electromagnetism concepts with structured problem-solving techniques for JEE and NEET aspirants.
Last Updated: July 2026
Question Source: JEE Main 2021 PYQ
Topic: Kinetic Theory Of Gases