What is the answer to the Adiabatic Process NEET 2026 question?
Option C, 124.5 J, is correct: the gas supplies this work by losing internal energy. The Adiabatic Process NEET 2026 question asks how much work one mole of an ideal monatomic gas does on its surroundings when it expands without exchanging heat and cools from 60 K to 50 K.
The source is NEET UG 2026, held on 21 June 2026, Code-50. Its medium difficulty is the question bank’s classification.
The given values are:
How do you write the first law with the correct work sign?
Use work done by the gas, not work done on it. With this convention, the internal-energy change equals the heat absorbed minus the work done by the gas:
For this adiabatic process, heat exchange is zero. Substitute that before using any numbers:
Zero heat exchange does not mean zero work or zero internal-energy change. The gas can use its internal energy to do work on the surroundings.
For an ideal gas, falling temperature means falling internal energy. The work done by the gas must therefore be positive:
How do you calculate the molar heat capacity from gamma?
Use the constant-volume molar heat capacity for the ideal gas’s internal-energy change. Derive its value from the given heat-capacity ratio using the molar specific-heat relations in NCERT Class 11 Thermodynamics:
The first relation gives:
Substitute into the second relation and collect terms:
Now evaluate the denominator and heat capacity:
This is a molar heat capacity, so retain the number of moles. The internal-energy equation is:
Using this heat capacity does not imply constant volume. Ideal-gas internal energy depends only on temperature, so this relation also applies during expansion.
How do you get 124.5 J from the temperature drop?
The temperature change is negative, giving an internal-energy loss of 124.5 J. The work done by the gas is the negative of that change, so the answer is positive.
Subtract the initial temperature from the final temperature:
Calculate the internal-energy change:
The units cancel correctly:
Apply the first-law result:
Choose option C. The physical check agrees: the gas cools as its internal energy supplies positive work during expansion.
These steps give the compact expression:
Pressure and volume are unnecessary because the temperature change already determines the internal-energy change. There is no need to use the following reversible ideal-gas relation, which does not apply to every adiabatic process:
Why does using the gas constant instead of heat capacity give option B?
Option B follows from replacing the molar heat capacity with the universal gas constant. Using the temperature-drop magnitude, that incorrect route gives:
NCERT’s relation gives the difference between the two molar heat capacities. It does not make the constant-volume heat capacity equal to the gas constant:
For this monatomic gas:
The 83 J calculation is missing this factor:
Both calculations have units of joules because the gas constant and molar heat capacity have the same units. Dimensional checking alone cannot catch this error: check the heat-capacity relation before substituting.
Can you apply the same method to three related questions?
The same first-law method handles compression, a different heat-capacity ratio and nonzero heat exchange. These are original practice variations, not additional verified NEET PYQs. Retain the same gas constant throughout:
Try each question before reading its calculation. Write the work convention first, then calculate the signed temperature change.
- One mole of the same monatomic gas is compressed adiabatically from 50 K to 60 K. Find work done by and on the gas.
The temperature rises, so internal energy increases. With no heat entering, work done on the gas supplies that increase.
Check: reversing the temperature change reverses the work sign for this adiabatic process. Keep “by” and “on” attached to your answers.
- One mole of an ideal gas cools adiabatically from 60 K to 50 K with the following heat-capacity ratio. Find work done by the gas.
Recalculate the heat capacity rather than carrying over the monatomic value. Then use the same internal-energy equation.
Check: changing gamma changes the molar heat capacity. The same temperature drop need not give the same work.
- One mole of the original monatomic gas cools from 60 K to 50 K while absorbing 40 J of heat. Find work done by the gas and decide whether the process is adiabatic.
The internal-energy change remains the same because the gas and temperature change are unchanged. Retain the heat term this time.
Check: this process is not adiabatic because heat exchange is nonzero. Before setting heat exchange to zero in your next question, underline the words that justify doing so.
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Frequently asked questions
What is the answer to the Adiabatic Process NEET 2026 question?
Option C, 124.5 J, is correct. For one mole of the ideal monatomic gas, Cv = 3R/2 = 12.45 J mol⁻¹ K⁻¹, so cooling from 60 K to 50 K gives an internal-energy change of −124.5 J. Since heat exchange is zero, the work done by the gas is +124.5 J.
Why is work done by the gas positive during adiabatic expansion?
With work defined as work done by the gas, the first law is ΔU = Q − W. In the given adiabatic expansion, Q = 0 and the gas cools, so ΔU is negative. Therefore, W = −ΔU is positive: the gas supplies work by losing internal energy.
Why do we use Cv in an adiabatic process if volume changes?
For an ideal gas, internal energy depends only on temperature, so ΔU = nCvΔT also applies when volume changes. Using Cv does not mean the process occurs at constant volume. Calculate the molar heat capacity from Cv = R/(γ − 1).
Why is 83 J the wrong answer to this adiabatic question?
The 83 J result comes from incorrectly using W = nR(T₁ − T₂). For this monatomic gas, Cv = 3R/2, not R, so that calculation misses a factor of 3/2. The correct work is 83 × 3/2 = 124.5 J.