In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (gamma=5/3) decreases from 60K to 50K. The work done by the gas in the process is : (Take the universal gas constant as R=8.3 J mol^-1 K^-1)

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q17 – NEET-UG 2026

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Home Physics ThermodynamicsQuestion #45

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (

γ = 5 / 3

) decreases from

60 K

50 K

. The work done by the gas in the process is : (Take the universal gas constant as

R = 8.3 {J mol}^{- 1} K^{- 1}

)

Clue 1 of 3: Adiabatic Process

An adiabatic process is one where no heat exchange occurs between the system and its surroundings.

🥷

Ninja Strategy•Factor Analysis

Calculate the numerator $n R (T_{1} - T_{2}) = 83 J$ and the denominator $(γ - 1) = 2 / 3$ . The work done must be $83 / (2 / 3) = 83 \times 1.5 = 124.5 J$ , quickly eliminating other options.

🧠 Full Solution Path

Step 1: Identify the Process and Formula✦ Active

The problem describes an adiabatic expansion of an ideal monatomic gas. The work done by the gas in an adiabatic process is given by the formula:

W = \frac{n R (T_{1} - T_{2})}{γ - 1}

Step 2: Determine Given Values○ Expand

We are given the following values: Number of moles, $n = 1 mole$ Universal gas constant, $R = 8.3 {J mol}^{- 1} K^{- 1}$ Initial temperature, $T_{1} = 60 K$ Final temperature, $T_{2} = 50 K$ For a monatomic gas, the adiabatic index, $γ = 5 / 3$ .

💡 Teacher's Secret Hint

Remember that $γ = C_{p} / C_{v}$ . For a monatomic gas, $C_{v} = \frac{3}{2} R$ and $C_{p} = \frac{5}{2} R$ , so $γ = 5 / 3$ .

Step 3: Calculate Work Done○ Expand

First, calculate $(γ - 1)$ :

γ - 1 = \frac{5}{3} - 1 = \frac{2}{3}

Now, substitute all the values into the work done formula:

W = \frac{1 \times 8.3 \times (60 - 50)}{2 / 3}

Simplify the expression:

W = \frac{8.3 \times 10}{2 / 3} = \frac{83}{2 / 3} = 83 \times \frac{3}{2}

Calculate the final value:

W = \frac{249}{2} = 124.5 J

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation. The work done is positive, indicating work done BY the gas during expansion.

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Physics Question 17 – NEET-UG 2026

Clue 1 of 3: Adiabatic Process

🧠 Full Solution Path