Question

A certain amount of gas of volume V at 27°C temperature and pressure 2 × 10⁷ Nm^–2 expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use, γ = 1.5)

• A. \(3.536\times10^5 \)Pa
• B. \(3.536\times10^6\) Pa
• C. \(1.25\times10^6\) Pa
• D. \(1.25\times10^5 \)Pa

Answer 1

The Correct Option is B

To solve this problem, we need to apply the concepts of isothermal and adiabatic processes for an ideal gas.

Initially, we have a gas with volume \( V \), temperature \( 27^\circ C \) (or 300 K), and pressure \( 2 \times 10^7 \, \text{Nm}^{-2} \).

Step 1: Isothermal Expansion

During an isothermal process, the temperature remains constant. According to Boyle's Law for an isothermal process:

P_1 V_1 = P_2 V_2

Initially, V_1 = V and P_1 = 2 \times 10^7 \, \text{Nm}^{-2}. The volume, V_2, after doubling is 2V.

We have:

2 \times 10^7 \times V = P_2 \times 2V

Solving for P_2:

P_2 = \frac{2 \times 10^7}{2} = 1 \times 10^7 \, \text{Nm}^{-2}

Step 2: Adiabatic Expansion

In an adiabatic process, no heat is exchanged. The equation for an adiabatic process is given by:

P_2 V_2^{\gamma} = P_3 V_3^{\gamma}

We have V_3 = 2V_2 = 4V and \gamma = 1.5.

Substituting the known values:

1 \times 10^7 \times (2V)^{1.5} = P_3 \times (4V)^{1.5}

Simplifying:

(2)^{1.5} = 2^{1.5} = \sqrt{8} = 2\sqrt{2}

P_3 = \frac{1 \times 10^7 \times 2^{1.5}}{4^{1.5}}

(4)^{1.5} = \sqrt{64} = 8

Thus:

P_3 = \frac{1 \times 10^7 \times 2\sqrt{2}}{8}

Calculating further:

P_3 = \frac{1 \times 10^7 \times 2\sqrt{2}}{8} = 3.536 \times 10^6 \, \text{Pa}

Thus, the final pressure of the gas is \(3.536 \times 10^6 \, \text{Pa}\), which is option B.