One mole of an ideal gas `(C_(v,m)=(5)/(2)R)` at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is :

Options

A. 270 K
B. 273 K
C. 248.5 K
D. 200 K

Explanation

To solve the problem of finding the final temperature of one mole of an ideal gas expanded adiabatically, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Number of moles (n) = 1 mole
- Specific heat at constant volume (C_v) = \( \frac{5}{2} R \)
- Initial temperature (T1) = 300 K
- Initial pressure (P1) = 5 atm
- Final pressure (P2) = 2 atm
- The process is adiabatic, meaning no heat is exchanged (Q = 0).

2. **Apply the First Law of Thermodynamics:**
For an adiabatic process, the first law of thermodynamics states:
\[ \Delta U = Q - W \]
Since \( Q = 0 \):
\[ \Delta U = -W \]

3. **Express Change in Internal Energy:**
The change in internal energy (\( \Delta U \)) can be expressed as:
\[ \Delta U = n C_v (T_2 - T_1) \]
Therefore, we have:
\[ n C_v (T_2 - T_1) = -W \]

4. **Calculate Work Done (W):**
The work done on the gas against a constant external pressure can be expressed as:
\[ W = P_{ext} \Delta V = P_{ext} (V_2 - V_1) \]
Since we are expanding against \( P_{ext} = P_2 = 2 \) atm, we can write:
\[ W = -P_2 (V_2 - V_1) \]

5. **Use the Ideal Gas Law to Relate Volumes:**
Using the ideal gas law \( PV = nRT \), we can express the volumes:
\[ V_1 = \frac{nRT_1}{P_1} \quad \text{and} \quad V_2 = \frac{nRT_2}{P_2} \]
Thus, the change in volume (\( \Delta V \)) can be expressed as:
\[ \Delta V = V_2 - V_1 = \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \]

6. **Substituting into the Work Equation:**
Substitute \( \Delta V \) into the work equation:
\[ W = -P_2 \left( \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \right) \]
This simplifies to:
\[ W = -nR(T_2 - \frac{P_2}{P_1} T_1) \]

7. **Substituting Work into the Internal Energy Equation:**
Now substituting \( W \) into the internal energy equation:
\[ n C_v (T_2 - T_1) = nR(T_2 - \frac{P_2}{P_1} T_1) \]

8. **Canceling n and Rearranging:**
Cancel \( n \) from both sides:
\[ C_v (T_2 - T_1) = R(T_2 - \frac{P_2}{P_1} T_1) \]
Substitute \( C_v = \frac{5}{2} R \):
\[ \frac{5}{2} R (T_2 - 300) = R(T_2 - \frac{2}{5} \cdot 300) \]

9. **Simplifying the Equation:**
Cancel \( R \) from both sides:
\[ \frac{5}{2} (T_2 - 300) = T_2 - 120 \]
Expanding gives:
\[ \frac{5}{2} T_2 - 750 = T_2 - 120 \]

10. **Rearranging to Solve for T2:**
Bringing like terms together:
\[ \frac{5}{2} T_2 - T_2 = 750 - 120 \]
This simplifies to:
\[ \frac{3}{2} T_2 = 630 \]
Therefore:
\[ T_2 = \frac{630 \times 2}{3} = 420 \text{ K} \]

### Final Answer:
The final temperature of the gas after adiabatic expansion is \( T_2 = 420 \text{ K} \).