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
Physics Expert of PadhoLikhoJEE
Specializes in breaking down complex mechanics and electromagnetism concepts with structured problem-solving techniques for JEE and NEET aspirants.
5 years of experience in online and content creation.
Last Updated: July 2026
Question Source: JEE Main 2021 PYQ
Topic: Kinetic Theory Of Gases