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Question

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

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$124.5\text{ J}$

Key Formula for Adiabatic Process

For an adiabatic process involving an ideal gas, the work done ($W$) is related to the change in internal energy ($\Delta U$). According to the first law of thermodynamics, $\Delta U = Q - W$. Since the process is adiabatic, $Q = 0$, hence $\Delta U = -W$.

The change in internal energy for $n$ moles of an ideal gas is given by:

$\Delta U = n C_V \Delta T$

Where $C_V$ is the molar specific heat at constant volume and $\Delta T$ is the change in temperature. For a monatomic ideal gas, $C_V = \frac{R}{\gamma - 1}$.

Combining these, the work done by the gas is:

$W = -\Delta U = -n C_V \Delta T = -n \left(\frac{R}{\gamma - 1}\right) (T_2 - T_1) = n \left(\frac{R}{\gamma - 1}\right) (T_1 - T_2)$

Step-by-Step Calculation

  • Identify the given values:
    • Number of moles, $n = 1$ mol
    • Adiabatic index, $\gamma = 5/3$
    • Initial temperature, $T_1 = 60$ K
    • Final temperature, $T_2 = 50$ K
    • Universal gas constant, $R = 8.3$ J mol-1 K-1
  • Calculate the term $(\gamma - 1)$:

    $\gamma - 1 = \frac{5}{3} - 1 = \frac{5 - 3}{3} = \frac{2}{3}$

  • Calculate the temperature difference $(T_1 - T_2)$:

    $T_1 - T_2 = 60 \text{ K} - 50 \text{ K} = 10 \text{ K}$

  • Substitute the values into the work done formula:

    $W = n \left(\frac{R}{\gamma - 1}\right) (T_1 - T_2)$

    $W = (1 \text{ mol}) \left(\frac{8.3 \text{ J mol}^{-1} \text{ K}^{-1}}{2/3}\right) (10 \text{ K})$

  • Simplify the expression:

    $W = 1 \times \left(8.3 \times \frac{3}{2}\right) \times 10 \text{ J}$

    $W = \frac{3}{2} \times 8.3 \times 10 \text{ J}$

    $W = 1.5 \times 8.3 \times 10 \text{ J}$

    $W = 12.45 \times 10 \text{ J}$

    $W = 124.5 \text{ J}$

  • Conclusion: The work done by the gas is $124.5$ J.
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  5. Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $32 \times 10^{18}$ /s then collision frequency in gas A is _________ /s.
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