With what minimum pressure (in kPa), a given volume of an ideal gas `(C_(p,m)=7//2R)` originally at 400 K and 100 kPa pressure can be compressed irreversibly adiabatically in order to raise its temperature to 600 K :

Options

A. 362.5 kPa
B. 275 kPa
C. 437.5 kPa
D. 550 kPa

Explanation

To solve the problem of determining the minimum pressure required for an ideal gas to be compressed irreversibly adiabatically from an initial state to a final state, we can follow these steps:

### Step 1: Understand the Given Information
- Initial temperature, \( T_1 = 400 \, K \)
- Final temperature, \( T_2 = 600 \, K \)
- Initial pressure, \( P_1 = 100 \, kPa \)
- Molar specific heat at constant pressure, \( C_{p,m} = \frac{7}{2} R \)

### Step 2: Calculate \( C_v \)
Using the relation \( C_p - C_v = R \):
\[ C_v = C_p - R = \frac{7}{2} R - R = \frac{5}{2} R \]

### Step 3: Apply the First Law of Thermodynamics
For an adiabatic process, the change in internal energy (\( \Delta U \)) is equal to the work done (\( W \)):
\[ \Delta U = W \]
Where:
\[ \Delta U = n C_v (T_2 - T_1) \]
And for work done in an irreversible adiabatic process:
\[ W = -P_{\text{external}} \Delta V \]

### Step 4: Relate Volume and Pressure
Using the ideal gas law:
\[ PV = nRT \]
We can express the volumes in terms of pressures and temperatures:
\[ V_1 = \frac{nRT_1}{P_1}, \quad V_2 = \frac{nRT_2}{P_2} \]
Thus,
\[ \Delta V = V_2 - V_1 = \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \]

### Step 5: Substitute into the First Law Equation
Substituting \( \Delta U \) and \( W \) into the equation:
\[ n C_v (T_2 - T_1) = -P_{\text{external}} \left( \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \right) \]
Cancel \( n \) from both sides:
\[ C_v (T_2 - T_1) = -P_{\text{external}} \left( \frac{RT_2}{P_2} - \frac{RT_1}{P_1} \right) \]

### Step 6: Substitute Known Values
Substituting \( C_v = \frac{5}{2} R \):
\[ \frac{5}{2} R (600 - 400) = -P_2 \left( \frac{RT_2}{P_2} - \frac{RT_1}{P_1} \right) \]
This simplifies to:
\[ \frac{5}{2} R (200) = -P_2 \left( \frac{R \cdot 600}{P_2} - \frac{R \cdot 400}{100} \right) \]

### Step 7: Simplify and Solve for \( P_2 \)
Cancelling \( R \) from both sides:
\[ 500 = -P_2 \left( \frac{600}{P_2} - 4 \right) \]
This leads to:
\[ 500 = -600 + 4P_2 \]
Rearranging gives:
\[ 4P_2 = 1100 \implies P_2 = \frac{1100}{4} = 275 \, kPa \]

### Final Answer
The minimum pressure required is:
\[ \boxed{275 \, kPa} \]