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Step-by-Step Solution
Step 1: Identify the Given Parameters
• Initial pressure of the gas, $P_1 = 2\ \text{atm}$
• Initial temperature of the gas, $T_1 = 27^\circ \text{C} = 300\ \text{K}$
• Final temperature of the gas, $T_2 = 927^\circ \text{C} = 1200\ \text{K}$
• Ratio of specific heats, $\gamma = 1.4$
Step 2: Recall the Adiabatic Relation
For an adiabatic process involving an ideal gas, the following relation holds:
$\displaystyle P^{\,1 - \gamma}\,T^{\,\gamma} = \text{constant}$
Step 3: Apply the Relation to Initial and Final States
Using the above formula for the initial state ($P_1, T_1$) and the final state ($P_2, T_2$):
$\displaystyle P_1^{\,1-\gamma}\,T_1^{\,\gamma} \;=\; P_2^{\,1-\gamma}\,T_2^{\,\gamma}$
Rearranging to solve for $\displaystyle \frac{P_2}{P_1}$, we get:
$\displaystyle \left(\frac{P_2}{P_1}\right)^{\,1-\gamma} \;=\; \left(\frac{T_2}{T_1}\right)^{\,\gamma}$
Step 4: Substitute Known Values
Substitute $\gamma = 1.4$, $T_1 = 300\ \text{K}$, and $T_2 = 1200\ \text{K}$:
$\displaystyle \left(\frac{P_2}{2\,\text{atm}}\right)^{\,1 - 1.4} \;=\; \left(\frac{1200}{300}\right)^{\,1.4}$
This simplifies to:
$\displaystyle \left(\frac{P_2}{2}\right)^{-\,0.4} \;=\; 4^{\,1.4}$
Step 5: Solve for $P_2$
Take the reciprocal exponent on both sides:
$\displaystyle \left(\frac{P_2}{2}\right)^{\,0.4} = 4^{\,1.4}$
Now, isolate $P_2$:
$\displaystyle \frac{P_2}{2} = \left( 4^{\,1.4} \right)^{\frac{1}{0.4}} \;=\; 4^{\frac{1.4}{0.4}} \;=\; 4^{3.5}$
We note that $4^{3.5} = (2^2)^{3.5} = 2^{7} = 128.$ Hence:
$\displaystyle \frac{P_2}{2} = 128 \quad \Longrightarrow \quad P_2 = 256\ \text{atm}$
Step 6: State the Final Answer
Therefore, the final pressure of the gas, $P_2$, is:
$\displaystyle \boxed{256\ \text{atm}}$
