Solution
1. System Definition & Approach
We analyze this problem using the First Law of Thermodynamics and the Work-Energy Theorem. Instead of using enthalpy directly, we define our thermodynamic system as the specific mass of gas (let’s say $n$ moles) that eventually fills the evacuated flask.
- Initial State: The gas packet is outside the flask at atmospheric pressure $P_0$ and temperature $T_0$. It occupies a volume $V_{in}$ equal to the flask’s volume (scaled by density, but conceptually the volume displaced).
- Process: The cork is removed. The atmosphere acts like a piston, pushing this gas packet into the vacuum. The flask is rigid and insulated ($Q = 0$).
- Final State: The gas is inside the flask at temperature $T_f$. The final kinetic energy of bulk motion is zero (gas comes to rest).
2. Applying the First Law
According to the First Law of Thermodynamics for this system:
$$ \Delta U = Q + W_{\text{on gas}} $$Since the flask is insulated and the process is rapid, it is adiabatic:
$$ Q = 0 $$ $$ \therefore \Delta U = W_{\text{on gas}} $$Calculating Work Done ($W_{\text{on gas}}$)
The atmosphere exerts a constant pressure $P_0$ on the boundary of our system, pushing it into the flask. The “piston” (atmosphere) moves through the volume originally occupied by the gas ($V_{in}$). The resisting pressure inside the vacuum is 0.
$$ W_{\text{on gas}} = P_0 \Delta V = P_0 (V_{\text{in}} – 0) = P_0 V_{\text{in}} $$Calculating Internal Energy Change ($\Delta U$)
For an ideal gas, the change in internal energy depends only on temperature:
$$ \Delta U = n C_V (T_f – T_0) $$3. Deriving the Final Temperature
Equating the work done to the change in internal energy:
$$ P_0 V_{\text{in}} = n C_V (T_f – T_0) $$From the Ideal Gas Law for the gas packet before entering ($P_0 V_{\text{in}} = n R T_0$), we substitute $P_0 V_{\text{in}}$:
$$ n R T_0 = n C_V (T_f – T_0) $$Canceling $n$ and rearranging:
$$ R T_0 = C_V T_f – C_V T_0 $$ $$ (R + C_V) T_0 = C_V T_f $$We know Mayer’s Relation: $C_P = C_V + R$. Substituting this on the left side:
$$ C_P T_0 = C_V T_f $$Solving for $T_f$:
$$ T_f = \frac{C_P}{C_V} T_0 = \gamma T_0 $$4. Numerical Calculation
The gas is mono-atomic, so the adiabatic index is $\gamma = \frac{5}{3}$.
Given initial temperature: $T_0 = 27^\circ\text{C} = 300 \text{ K}$.
$$ T_f = \frac{5}{3} \times 300 \text{ K} = 500 \text{ K} $$Converting back to Celsius:
$$ T_f (^\circ\text{C}) = 500 – 273 = 227^\circ\text{C} $$Correct Option: (d)