Two moles of an ideal gas have an initial kelvin temperature `T_i` and absolute pressure `P_i` the gas undergoes a reversible isothermal compression from an initial volume `V_i` to a final volume 0.5 `V_i`
The ratio of the molar specific heat capacity at constant pressure to that at constant volume, `gamma `,for diatomic hydrogen gas is 7/5. In an adiabatic compression, the gas, originally at atmospheric pressure, is compressed from an original volume of `0.30 m^3 ` to `0.15 m^3 ` What is the final pressure of the gas?

Options

A. `2.0 xx 10^5 Pa`
B. `2.7 xx 10^5 Pa`
C. `3.0 xx 10^5 Pa`
D. `3.7 xx 10^5 Pa`

Answer

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: Identify the given data
- Initial pressure, \( P_1 = 1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa} \)
- Initial volume, \( V_1 = 0.30 \, \text{m}^3 \)
- Final volume, \( V_2 = 0.15 \, \text{m}^3 \)
- The ratio of specific heats, \( \gamma = \frac{C_p}{C_v} = \frac{7}{5} = 1.4 \)

### Step 2: Use the adiabatic process equation
For an adiabatic process, the relationship between pressure and volume can be expressed as:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
From this, we can derive the final pressure \( P_2 \):
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ P_2 = 1.01 \times 10^5 \, \text{Pa} \left( \frac{0.30 \, \text{m}^3}{0.15 \, \text{m}^3} \right)^{1.4} \]

### Step 4: Calculate the volume ratio
Calculate the volume ratio:
\[ \frac{V_1}{V_2} = \frac{0.30}{0.15} = 2 \]

### Step 5: Raise the volume ratio to the power of gamma
Now raise the volume ratio to the power of \( \gamma \):
\[ 2^{1.4} \approx 2.639 \]

### Step 6: Calculate the final pressure
Now substitute this back into the equation for \( P_2 \):
\[ P_2 = 1.01 \times 10^5 \, \text{Pa} \times 2.639 \]
\[ P_2 \approx 2.66 \times 10^5 \, \text{Pa} \]

### Step 7: Final Answer
Thus, the final pressure \( P_2 \) of the gas after adiabatic compression is approximately:
\[ P_2 \approx 2.66 \times 10^5 \, \text{Pa} \]

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