Internal Energy Decrease in Monatomic Ideal Gas Adiabatic Expansion

To find the magnitude of the decrease in the internal energy of the gas, we proceed with the following steps:

1. Determine the adiabatic index ($\gamma$) and heat capacity ($C_v$) for a monatomic ideal gas: For a monatomic ideal gas, the degrees of freedom $f = 3$. The molar heat capacity at constant volume is: $C_v = \frac{f}{2}R = \frac{3}{2}R$ Given $R = 8.0\text{ J mol}^{-1}\text{K}^{-1}$, we have: $C_v = \frac{3}{2} \times 8.0 = 12.0\text{ J mol}^{-1}\text{K}^{-1}$

   The adiabatic index $\gamma$ is given by: $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$

2. Calculate the final temperature ($T_2$): For an adiabatic process involving an ideal gas, the relationship between temperature and volume is given by: $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

   Substituting $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$ and $V_2 = 8V_1$: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}$ $T_2 = 100 \times \left( \frac{1}{8} \right)^{2/3}$ Since $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$: $T_2 = \frac{100}{4} = 25\text{ K}$

3. Calculate the decrease in internal energy ($-\Delta U$): The change in internal energy for $n = 1\text{ mole}$ of an ideal gas is: $\Delta U = n C_v (T_2 - T_1)$ $\Delta U = 1 \times 12.0 \times (25 - 100) = 12.0 \times (-75) = -900\text{ J}$

Thus, the magnitude of the decrease in internal energy is: $-\Delta U = 900\text{ J}$

Final Answer: 900