1. Given data

• Ideal gas with molar heat capacity: $\bar C_v = \frac{5}{2}R$
• Adiabatic process: $q=0$
• External pressure is constant: $P_{\text{ext}}=1\,\text{atm}$
• Initial state: $T_1=298\,\text{K},\quad P_1=5\,\text{atm}$
• Final volume is double the initial volume: $V_2=2V_1$

We need the final temperature $T_2$.

2. Apply first law of thermodynamics

For an adiabatic process: $\Delta U = q+w = w$

For expansion against constant external pressure: $w=-P_{\text{ext}}(V_2-V_1)$

For an ideal gas: $\Delta U = n\bar C_v(T_2-T_1)$

So, $n\bar C_v(T_2-T_1)=-P_{\text{ext}}(V_2-V_1)$

Since $V_2=2V_1$, $V_2-V_1=V_1$

Hence, $n\bar C_v(T_2-T_1)=-P_{\text{ext}}V_1$

3. Use ideal gas equation for initial state

Initially, $P_1V_1=nRT_1$

Thus, $V_1=\frac{nRT_1}{P_1}$

Substitute into energy equation: $n\bar C_v(T_2-T_1)=-P_{\text{ext}}\left(\frac{nRT_1}{P_1}\right)$

Cancel $n$: $\bar C_v(T_2-T_1)=-\frac{P_{\text{ext}}}{P_1}RT_1$

Now substitute $\bar C_v=\frac{5}{2}R$: $\frac{5}{2}R(T_2-T_1)=-\frac{1}{5}RT_1$

Cancel $R$: $\frac{5}{2}(T_2-T_1)=-\frac{T_1}{5}$

With $T_1=298\,\text{K}$: $\frac{5}{2}(T_2-298)=-\frac{298}{5}$

So, $T_2-298=-\frac{298}{5}\cdot\frac{2}{5}=-\frac{596}{25}=-23.84$

Therefore, $T_2=298-23.84=274.16\,\text{K}$

Nearest integer: $\boxed{274\,\text{K}}$

4. Comparison with stored answer

Stored correct answer = 274

Our derived answer = 274

So the answer agrees.