© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Quantities
• Number of moles of the monoatomic gas, $n = 2$.
• Initial volume, $V_1 = V$.
• Final volume, $V_2 = 2V$.
• Initial temperature, $T_1 = 27^\circ \text{C} = 300\,\text{K}$.
• For a monoatomic gas, degrees of freedom $f = 3$.
Step 2: Determine the Adiabatic Exponent (γ)
The ratio of specific heats, $ \gamma = \dfrac{C_p}{C_v} $. For a monoatomic gas (with $f = 3$):
\[
\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{3} = \dfrac{5}{3}.
\]
Step 3: Use the Adiabatic Relation to Find the Final Temperature
In an adiabatic process for an ideal gas,
\[
T\,V^{\,\gamma - 1} = \text{constant}.
\]
Hence,
\[
T_1 \, V_1^{\,\gamma - 1} = T_2 \, V_2^{\,\gamma - 1}.
\]
Substituting $T_1 = 300\,\text{K}$, $V_1 = V$, and $V_2 = 2V$, we have:
\[
300 \, (V)^{\,\gamma - 1} = T_2 \,(2V)^{\,\gamma - 1}.
\]
Since $\gamma = \tfrac{5}{3}$, then $\gamma - 1 = \tfrac{2}{3}$. Therefore,
\[
300 = T_2 \,\left(\dfrac{2V}{V}\right)^{\tfrac{2}{3}}
= T_2 \,\left(2\right)^{\tfrac{2}{3}}.
\]
Thus,
\[
T_2 = \dfrac{300}{(2)^{\tfrac{2}{3}}} \approx 189\,\text{K}.
\]
Step 4: Calculate the Change in Internal Energy (ΔU)
For an ideal gas, the change in internal energy is given by:
\[
\Delta U = \dfrac{f}{2}\,nR\,\Delta T
= \dfrac{f}{2}\,nR\,(T_2 - T_1).
\]
Here, $f = 3$, $n = 2$, and $R \approx 8.31\,\text{J mol}^{-1}\text{K}^{-1}$. Substituting values,
\[
\Delta U
= \dfrac{3}{2} \times 2 \times 8.31 \times (189 - 300)\,\text{J}.
\]
\[
= 3 \times 8.31 \times ( -111 )\,\text{J}
= -2763.3\,\text{J} \approx -2.76 \times 10^3\,\text{J}.
\]
Converting to kJ,
\[
\Delta U \approx -2.7\,\text{kJ}.
\]
Step 5: Final Answer
(a) Final Temperature, $T_2 = 189\,\text{K}$.
(b) Change in Internal Energy, $\Delta U \approx -2.7\,\text{kJ}$.
