One mole of gas of specific heat ratio 1.5 being initially at temperature 290 K is adiabatically compressed to increase its pressure 8 times. The temperature of the gas after compression will be

Options

A. 580 K
B. 870 K
C. `270sqrt2` K
D. 1160 K

Answer

To solve the problem of finding the temperature of a gas after adiabatic compression, we can follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure (P), volume (V), and temperature (T) for an ideal gas undergoing an adiabatic process is given by the equation:
\[ P V^\gamma = \text{constant} \]
where \( \gamma \) (gamma) is the specific heat ratio.

### Step 2: Define Initial and Final Conditions
Given:
- Initial temperature \( T_1 = 290 \, \text{K} \)
- Specific heat ratio \( \gamma = 1.5 \)
- The pressure increases 8 times, so:
- Initial pressure \( P_1 = P \)
- Final pressure \( P_f = 8P_1 \)

### Step 3: Apply the Adiabatic Condition
From the adiabatic condition, we can write:
\[ P_1 V_1^\gamma = P_f V_f^\gamma \]
Substituting \( P_f = 8P_1 \):
\[ P_1 V_1^\gamma = 8P_1 V_f^\gamma \]

### Step 4: Cancel Out \( P_1 \)
Since \( P_1 \) is common on both sides, we can cancel it:
\[ V_1^\gamma = 8 V_f^\gamma \]

### Step 5: Relate Volume to Temperature
Using the ideal gas law, we can express volume in terms of temperature and pressure:
\[ PV = nRT \]
Thus, we can express volume as:
\[ V = \frac{nRT}{P} \]
Substituting this into our equation gives us:
\[ \left(\frac{nRT_1}{P_1}\right)^\gamma = 8 \left(\frac{nRT_2}{8P_1}\right)^\gamma \]

### Step 6: Simplify the Equation
This simplifies to:
\[ \left(\frac{T_1}{P_1}\right)^\gamma = 8 \left(\frac{T_2}{8P_1}\right)^\gamma \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ \frac{T_1^\gamma}{P_1^\gamma} = \frac{T_2^\gamma}{8^{\gamma-1} P_1^\gamma} \]
Cancelling \( P_1^\gamma \) from both sides:
\[ T_1^\gamma = T_2^\gamma \cdot 8^{\gamma - 1} \]

### Step 8: Substitute Known Values
Substituting \( \gamma = 1.5 \) and \( T_1 = 290 \, \text{K} \):
\[ 290^{1.5} = T_2^{1.5} \cdot 8^{0.5} \]

### Step 9: Solve for \( T_2 \)
Calculating \( 8^{0.5} = 2 \):
\[ 290^{1.5} = 2 T_2^{1.5} \]
Now, rearranging gives:
\[ T_2^{1.5} = \frac{290^{1.5}}{2} \]

### Step 10: Take the Cube Root
To find \( T_2 \):
\[ T_2 = \left(\frac{290^{1.5}}{2}\right)^{\frac{2}{3}} \]

### Step 11: Calculate \( T_2 \)
Calculating \( 290^{1.5} \):
\[ 290^{1.5} = 290 \times \sqrt{290} \approx 290 \times 17.029 = 4934.41 \]
Now substituting:
\[ T_2 = \left(\frac{4934.41}{2}\right)^{\frac{2}{3}} \approx (2467.205)^{\frac{2}{3}} \]

### Final Calculation
Calculating this gives:
\[ T_2 \approx 580 \, \text{K} \]

Thus, the final temperature of the gas after adiabatic compression is:
\[ \boxed{580 \, \text{K}} \]