Question

Initially the pressure of 1 mole of an ideal gas is \( 10^5 \) Nm\(^2\) and its volume is 16 liters. When it is adiabatically compressed, its final volume is 2 liters. Work done on the gas is (molar specific heat at constant volume \( C_V = \frac{3R}{2} \)):

• A. \( 72 \) kJ
• B. \( 7.2 \) kJ
• C. \( 720 \) kJ
• D. \( 360 \) kJ

Hint:

For an adiabatic process, use the relation \( W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \). Ensure volume is in cubic meters before calculation.

Answer 1

The Correct Option is B

Step 1: Recall the formula for work done in an adiabatic process The work done in an adiabatic process is given by: \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \] Where: - \(P_1 = 10^5 \text{ N/m}^2\) - \(V_1 = 16 \text{ L} = 16 \times 10^{-3} \text{ m}^3\) - \(V_2 = 2 \text{ L} = 2 \times 10^{-3} \text{ m}^3\) The ratio of specific heats is: \[ \gamma = \frac{C_p}{C_v} = \frac{5R/2}{3R/2} = \frac{5}{3} \]
Step 2: Finding the final pressure using the adiabatic condition From the adiabatic relation: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] \[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \] \[ P_2 = 10^5 \left( \frac{16}{2} \right)^{5/3} \] \[ P_2 = 10^5 \times 8^{5/3} \] Since \( 8^{5/3} \approx 32 \), \[ P_2 \approx 32 \times 10^5 = 3.2 \times 10^6 \text{ N/m}^2 \]
Step 3: Calculate the Work Done \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \] \[ W = \frac{(10^5)(16 \times 10^{-3}) - (3.2 \times 10^6)(2 \times 10^{-3})}{\frac{5}{3} - 1} \] \[ W = \frac{(1600) - (6400)}{\frac{5}{3} - 1} \] \[ W = \frac{-4800}{\frac{2}{3}} = -4800 \times \frac{3}{2} = -7200 \text{ J} = -7.2 \text{ kJ} \] (Note: The negative sign indicates work is done *on* the gas.)

Answer 2

To solve this problem, we'll use the adiabatic process formula for an ideal gas, which relates pressure, volume, and the heat capacity ratio \(\gamma\). The work done \(W\) during an adiabatic process is given by:

\[ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \]

For an adiabatic process, the relation between pressure, volume, and \(\gamma\) can be represented as:

\( P_1V_1^\gamma = P_2V_2^\gamma \)

Given:

• Initial pressure \(P_1 = 10^5 \, \text{N/m}^2\)
• Initial volume \(V_1 = 16 \, \text{L} = 0.016 \, \text{m}^3\)
• Final volume \(V_2 = 2 \, \text{L} = 0.002 \, \text{m}^3\)
• \(C_V = \frac{3R}{2}\), hence \(\gamma = \frac{C_P}{C_V} = \frac{5}{3}\)

First, we need to find the final pressure \(P_2\) using the adiabatic relation:

\( P_1V_1^\gamma = P_2V_2^\gamma \)

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

Substituting the given values:

\( P_2 = 10^5 \left(\frac{0.016}{0.002}\right)^{\frac{5}{3}} \)

\( P_2 = 10^5 \times (8)^{\frac{5}{3}} \approx 10^5 \times 32 \)

\( P_2 \approx 3.2 \times 10^6 \, \text{N/m}^2 \)

Now, calculate the work done using the formula:

\( W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \)

\( W = \frac{10^5 \times 0.016 - 3.2 \times 10^6 \times 0.002}{\frac{5}{3} - 1} \)

\( W = \frac{1600 - 6400}{\frac{2}{3}} \)

\( W = -\frac{4800}{\frac{2}{3}} \)

\( W = -4800 \times \frac{3}{2} \)

\( W = -7200 \, \text{J} \)

The work done on the gas is 7200 J, or \(7.2 \, \text{kJ}\).

Hence, the correct answer is \(7.2 \, \text{kJ}\).