A monatomic gas initially at `17^@ C` has suddenly compressed adiabatically to one-eighth of its original volume. The temperature after compression is

Options

A. 887 K
B. 36.25 K
C. 2320 K
D. 1160 K

Answer

To solve the problem of finding the temperature after the adiabatic compression of a monatomic gas, we can use the following steps:

### Step 1: Understand the Initial Conditions
The initial temperature \( T_1 \) of the gas is given as \( 17^\circ C \). We need to convert this to Kelvin for calculations:
\[ T_1 = 17 + 273 = 290 \, K \]

**Hint:** Always convert Celsius to Kelvin by adding 273.

### Step 2: Use the Adiabatic Process Formula
For an adiabatic process involving an ideal gas, the relationship between the initial and final temperatures and volumes can be expressed as:
\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
where \( \gamma \) (gamma) is the heat capacity ratio, which for a monatomic gas is \( \frac{5}{3} \).

**Hint:** Remember that for a monatomic gas, \( \gamma = \frac{5}{3} \).

### Step 3: Determine the Volume Ratio
The gas is compressed to one-eighth of its original volume, so:
\[ V_2 = \frac{V_1}{8} \]
Thus, the volume ratio \( \frac{V_1}{V_2} \) is:
\[ \frac{V_1}{V_2} = 8 \]

**Hint:** When the volume is compressed to one-eighth, the volume ratio is the inverse, which is 8.

### Step 4: Substitute Values into the Formula
Now we can substitute \( T_1 \), \( \frac{V_1}{V_2} \), and \( \gamma \) into the adiabatic process formula:
\[ \frac{T_2}{290} = 8^{\frac{5}{3} - 1} \]
Calculating \( \frac{5}{3} - 1 \):
\[ \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]
So we have:
\[ \frac{T_2}{290} = 8^{\frac{2}{3}} \]

**Hint:** Simplify the exponent before calculating the power.

### Step 5: Calculate \( 8^{\frac{2}{3}} \)
We know \( 8 = 2^3 \), thus:
\[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \]
Now substituting back:
\[ \frac{T_2}{290} = 4 \]

**Hint:** Use properties of exponents to simplify calculations.

### Step 6: Solve for \( T_2 \)
Now we can solve for \( T_2 \):
\[ T_2 = 4 \times 290 = 1160 \, K \]

**Hint:** Multiply carefully to find the final temperature.

### Final Answer
The temperature after compression is:
\[ T_2 = 1160 \, K \]