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 follow these steps:

### Step 1: Convert the initial temperature to Kelvin
The initial temperature given is \(17^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_1 = 17 + 273 = 290 \, K \]

### Step 2: Identify the value of gamma (\(\gamma\))
For a monatomic gas, the value of \(\gamma\) (the ratio of specific heats) is given by:
\[ \gamma = \frac{C_p}{C_v} = \frac{5R/2}{3R/2} = \frac{5}{3} \]

### Step 3: Use the adiabatic process relation
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Given that the gas is compressed to one-eighth of its original volume, we have:
\[ V_2 = \frac{V_1}{8} \]
Substituting this into the equation gives:
\[ T_1 V_1^{\gamma - 1} = T_2 \left(\frac{V_1}{8}\right)^{\gamma - 1} \]

### Step 4: Simplify the equation
We can cancel \(V_1^{\gamma - 1}\) from both sides:
\[ T_1 = T_2 \left(\frac{1}{8}\right)^{\gamma - 1} \]
Rearranging gives:
\[ T_2 = T_1 \left(8\right)^{\gamma - 1} \]

### Step 5: Substitute the values
Now substituting the known values:
\[ T_2 = 290 \left(8\right)^{\frac{5}{3} - 1} \]
Calculating \(\gamma - 1\):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
Thus,
\[ T_2 = 290 \left(8\right)^{\frac{2}{3}} \]

### Step 6: Calculate \(8^{\frac{2}{3}}\)
Calculating \(8^{\frac{2}{3}}\):
\[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \]
So,
\[ T_2 = 290 \times 4 = 1160 \, K \]

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