An ideal gas at pressure `4 xx 10^(5)Pa` and temperature `400K` occupies `100cc`. It is adiabatically expanded to double of its original volume. Calculate (a) the final pressure (b) final temperature and (c ) work done by the gas in the process `(gamma = 1.5)`

Explanation

To solve the problem step by step, we will calculate the final pressure, final temperature, and work done by the gas during the adiabatic expansion.

### Given Data:
- Initial Pressure, \( P_1 = 4 \times 10^5 \) Pa
- Initial Temperature, \( T_1 = 400 \) K
- Initial Volume, \( V_1 = 100 \) cc = \( 100 \times 10^{-6} \) m³
- Final Volume, \( V_2 = 2V_1 = 200 \times 10^{-6} \) m³
- \( \gamma = 1.5 \)

### (a) Final Pressure \( P_2 \)

For an adiabatic process, the relationship between pressure and volume is given by:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Rearranging this equation to find \( P_2 \):

\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

Substituting the known values:

\[ P_2 = 4 \times 10^5 \left( \frac{100 \times 10^{-6}}{200 \times 10^{-6}} \right)^{1.5} \]

Calculating the fraction:

\[ P_2 = 4 \times 10^5 \left( \frac{1}{2} \right)^{1.5} \]

Calculating \( \left( \frac{1}{2} \right)^{1.5} \):

\[ \left( \frac{1}{2} \right)^{1.5} = \frac{1}{\sqrt{2^3}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]

Now substituting back:

\[ P_2 = 4 \times 10^5 \times \frac{\sqrt{2}}{4} = 10^5 \sqrt{2} \text{ Pa} \]

### (b) Final Temperature \( T_2 \)

For an adiabatic process, the relationship between temperature and volume is given by:

\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

Rearranging to find \( T_2 \):

\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

Substituting the known values:

\[ T_2 = 400 \left( \frac{100 \times 10^{-6}}{200 \times 10^{-6}} \right)^{1.5 - 1} \]

Calculating the fraction:

\[ T_2 = 400 \left( \frac{1}{2} \right)^{0.5} \]

Calculating \( \left( \frac{1}{2} \right)^{0.5} \):

\[ \left( \frac{1}{2} \right)^{0.5} = \frac{1}{\sqrt{2}} \]

Now substituting back:

\[ T_2 = 400 \times \frac{1}{\sqrt{2}} = 200\sqrt{2} \text{ K} \]

### (c) Work Done \( W \)

The work done in an adiabatic process can be calculated using the formula:

\[ W = \frac{nR \Delta T}{\gamma - 1} \]

Where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( \Delta T = T_2 - T_1 \).

First, we need to find \( n \) using the ideal gas equation:

\[ n = \frac{P_1 V_1}{R T_1} \]

Using \( R = 8.314 \, \text{J/(mol K)} \):

\[ n = \frac{(4 \times 10^5) \times (100 \times 10^{-6})}{8.314 \times 400} \]

Calculating \( n \):

\[ n = \frac{4 \times 10^5 \times 10^{-4}}{3325.6} \approx 0.120 \text{ moles} \]

Now, calculate \( \Delta T \):

\[ \Delta T = T_2 - T_1 = 200\sqrt{2} - 400 \]

Substituting \( n \) and \( \Delta T \) into the work done formula:

\[ W = \frac{0.120 \times 8.314 \times (200\sqrt{2} - 400)}{1.5 - 1} \]

Calculating \( W \):

\[ W = \frac{0.120 \times 8.314 \times (200\sqrt{2} - 400)}{0.5} \]

After calculating the above expression, we find:

\[ W \approx 42 - 42\sqrt{2} \text{ J} \]

### Summary of Results:
- (a) Final Pressure \( P_2 \approx 10^5 \sqrt{2} \) Pa
- (b) Final Temperature \( T_2 \approx 200\sqrt{2} \) K
- (c) Work Done \( W \approx 42 - 42\sqrt{2} \) J