Question

One mole of an ideal monoatomic gas is taken round the cyclic process MNOM. The work done by the gas is

• A. $4.5 P_0V_0 $
• B. $4 P_0V_0 $
• C. $9 P_0V_0 $
• D. $2 P_0V_0 $

Answer 1

The Correct Option is D

We are given the cyclic process MNOM for one mole of an ideal monoatomic gas. The graph indicates that the process involves the following steps: 
From \( O \) to \( N \), the gas undergoes an isobaric (constant pressure) expansion.
From \( N \) to \( M \), the gas undergoes an isochoric (constant volume) process.
From \( M \) to \( O \), the gas undergoes an isothermal (constant temperature) compression.
We need to calculate the total work done by the gas during the entire cyclic process.
Step 1: Work done during the isobaric process (O to N) In the isobaric process, the pressure is constant. The work done by the gas is given by: \[ W_{ON} = P \Delta V = P_0 (V_1 - V_0) \] where \( V_1 = 3V_0 \) (from the graph) and \( V_0 \) is the initial volume. Thus, the work done during the process from \( O \) to \( N \) is: \[ W_{ON} = P_0 (3V_0 - V_0) = 2 P_0 V_0 \] Step 2: Work done during the isochoric process (N to M) During the isochoric process, the volume is constant, so no work is done: \[ W_{NM} = 0 \] Step 3: Work done during the isothermal process (M to O) In the isothermal process, the temperature remains constant. The work done by the gas is given by the area under the curve on the pressure-volume graph. Since the process is along a hyperbola, the work done can be calculated using the formula for an isothermal process: \[ W_{MO} = nRT \ln \left( \frac{V_0}{V_1} \right) \] But this part does not contribute to the net work as the pressure and volume return to their initial values at the end of the cycle. Step 4: Total work done The total work done by the gas is the sum of the work done in each step of the cycle: \[ W_{\text{total}} = W_{ON} + W_{NM} + W_{MO} \] Since \( W_{NM} = 0 \) and \( W_{MO} = 0 \), the total work done by the gas is: \[ W_{\text{total}} = W_{ON} = 2 P_0 V_0 \] Thus, the work done by the gas in the cyclic process is \( 2 P_0 V_0 \).

Therefore, the correct answer is (D): \( 2 P_0 V_0 \).

Answer 2

The given process is a cyclic process, and the work done by the gas in one cycle is the area enclosed by the path on the PV diagram. The area of a triangle can be calculated as: \[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \] Looking at the given PV diagram, the base of the triangle is the change in volume, \( \Delta V = 3V_0 - V_0 = 2V_0 \), and the height is the change in pressure, \( \Delta P = 3P_0 - P_0 = 2P_0 \). Thus, the work done by the gas during the cyclic process is: \[ W = \frac{1}{2} \times \Delta V \times \Delta P = \frac{1}{2} \times 2V_0 \times 2P_0 \] \[ W = 2P_0 V_0 \] Thus, the work done by the gas is \( 2P_0 V_0 \).