Question

A monatomic gas at a pressure of 100 kPa expands adiabatically such that its final volume becomes 8 times its initial volume. If the work done during the process is 180 J, then the initial volume of the gas is

• A. 1600 cm\(^3\)
• B. 800 cm\(^3\)
• C. 1200 cm\(^3\)
• D. 2000 cm\(^3\)

Hint:

In adiabatic processes, remember to use the formula for work with correct value of \( \gamma \). For monatomic gases, \( \gamma = \frac{5}{3} \). Convert volume units carefully.

Answer 1

The Correct Option is A

Step 1: Use the formula for adiabatic work done: \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \] For a monatomic gas, \( \gamma = \frac{5}{3} \). Since it's an adiabatic process: \[ W = \frac{P_1 V_1}{\gamma - 1} \left[ 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right] \] Step 2: Given: \[ P_1 = 100 \times 10^3 \, \text{Pa},
V_2 = 8 V_1,
W = 180 \, \text{J} \] Substitute values: \[ 180 = \frac{100 \times 10^3 . V_1}{5/3 - 1} \left[ 1 - \left( \frac{1}{8} \right)^{2/3} \right] \] \[ \Rightarrow 180 = \frac{100000 . V_1}{2/3} \left[ 1 - \left( \frac{1}{8} \right)^{2/3} \right] \] Step 3: Compute numerically: \[ \left( \frac{1}{8} \right)^{2/3} = \left( 2^{-3} \right)^{2/3} = 2^{-2} = \frac{1}{4} \] \[ \Rightarrow 180 = \frac{100000 . V_1}{2/3} \left( 1 - \frac{1}{4} \right) = \frac{100000 . V_1}{2/3} . \frac{3}{4} \] \[ \Rightarrow 180 = 100000 . V_1 . \frac{3}{2} . \frac{3}{4} = 100000 . V_1 . \frac{9}{8} \] Step 4: Solve for \( V_1 \): \[ V_1 = \frac{180 . 8}{100000 . 9} = \frac{1440}{900000} = 0.0016 \, \text{m}^3 = 1600 \, \text{cm}^3 \] \[ \boxed{V_1 = 1600 \, \text{cm}^3} \]