A heat insulating cylinder with a movable piston contins 5 moles of hydrogen at standard temperature and prssure if the gas is cmpressed to quarter of its original volume then the pressure of the gas is increased by `(gamma=1.4)`

Options

A. `(2)^(1.4)`
B. `(3)^(1.4)`
C. `(4)^(1.4)`
D. `(5)^(1.4)`

Explanation

To solve the problem step by step, we will use the principles of thermodynamics, specifically focusing on the adiabatic process for an ideal gas.

### Step 1: Understand the Given Information
We have:
- Number of moles of hydrogen gas, \( n = 5 \) moles
- Initial volume, \( V_1 \)
- Final volume, \( V_2 = \frac{V_1}{4} \)
- The process is adiabatic (no heat exchange), so \( \Delta Q = 0 \)
- The adiabatic condition is described by the equation \( PV^\gamma = \text{constant} \), where \( \gamma = 1.4 \).

### Step 2: Apply the Adiabatic Condition
From the adiabatic condition, we can write:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

### Step 3: Substitute \( V_2 \)
Since \( V_2 = \frac{V_1}{4} \), we can substitute this into the equation:
\[ P_1 V_1^\gamma = P_2 \left(\frac{V_1}{4}\right)^\gamma \]

### Step 4: Simplify the Equation
This can be simplified to:
\[ P_1 V_1^\gamma = P_2 \frac{V_1^\gamma}{4^\gamma} \]
Now, we can cancel \( V_1^\gamma \) from both sides (assuming \( V_1 \neq 0 \)):
\[ P_1 = P_2 \frac{1}{4^\gamma} \]

### Step 5: Rearranging for \( P_2 \)
Rearranging the equation gives:
\[ P_2 = P_1 \cdot 4^\gamma \]

### Step 6: Calculate the Ratio of Pressures
To find the increase in pressure, we need to calculate the ratio \( \frac{P_2}{P_1} \):
\[ \frac{P_2}{P_1} = 4^\gamma \]

### Step 7: Substitute the Value of \( \gamma \)
Substituting \( \gamma = 1.4 \):
\[ \frac{P_2}{P_1} = 4^{1.4} \]

### Step 8: Calculate \( 4^{1.4} \)
Now we can calculate \( 4^{1.4} \):
\[ 4^{1.4} \approx 4^{1.4} = 4^{\frac{14}{10}} = (2^2)^{\frac{14}{10}} = 2^{\frac{28}{10}} = 2^{2.8} \]
Using a calculator or logarithmic tables, we can find that:
\[ 4^{1.4} \approx 7.211 \]

### Conclusion
Thus, the pressure of the gas after compression is approximately \( 7.211 \) times the initial pressure.

---