A gass of given mass at a pressure of `10^(5)Nm^(-2)` expands isothermally until its volume is doubled and then adiabatically until volume is again double. Find the final pressure of the gas. `(gamma=1.4)`

Answer

To solve the problem step by step, we will analyze the two processes the gas undergoes: isothermal expansion and adiabatic expansion.

### Step 1: Isothermal Expansion
In the isothermal process, the pressure and volume of the gas are related by the equation:
\[ P_1 V_1 = P_2 V_2 \]
where:
- \( P_1 = 10^5 \, \text{N/m}^2 \) (initial pressure)
- \( V_2 = 2V_1 \) (the volume doubles)

Substituting \( V_2 \) into the equation gives:
\[ P_1 V_1 = P_2 (2V_1) \]

We can cancel \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[ P_1 = 2P_2 \]

Now, solving for \( P_2 \):
\[ P_2 = \frac{P_1}{2} = \frac{10^5}{2} = 5 \times 10^4 \, \text{N/m}^2 \]

### Step 2: Adiabatic Expansion
In the adiabatic process, the relationship between pressure and volume is given by:
\[ P_2 V_2^\gamma = P_3 V_3^\gamma \]
where:
- \( \gamma = 1.4 \) (given)
- \( V_3 = 2V_2 = 4V_1 \) (the volume doubles again)

Substituting \( V_2 \) and \( V_3 \) into the equation gives:
\[ P_2 (2V_1)^\gamma = P_3 (4V_1)^\gamma \]

We can cancel \( V_1^\gamma \) from both sides:
\[ P_2 (2^\gamma) = P_3 (4^\gamma) \]

Now, we can express \( 4^\gamma \) as \( (2^2)^\gamma = 2^{2\gamma} \):
\[ P_2 (2^\gamma) = P_3 (2^{2\gamma}) \]

Dividing both sides by \( 2^\gamma \):
\[ P_3 = \frac{P_2}{2^\gamma} \]

Substituting \( P_2 = 5 \times 10^4 \, \text{N/m}^2 \):
\[ P_3 = \frac{5 \times 10^4}{2^{1.4}} \]

Calculating \( 2^{1.4} \):
\[ 2^{1.4} \approx 2.639 \]

Now substituting this value:
\[ P_3 \approx \frac{5 \times 10^4}{2.639} \approx 1.895 \times 10^4 \, \text{N/m}^2 \]

### Final Answer
Thus, the final pressure of the gas after both processes is approximately:
\[ P_3 \approx 1.89 \times 10^4 \, \text{N/m}^2 \]

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