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.