Question

One mole of an ideal monoatomic gas is subjected to changes as shown in the graph The magnitude of the work done (by the system or on the system) is _______ J (nearest integer)

Given : \(\log 2=0.3 \ln 10=2.3\)

Hint:

For work done in thermodynamic processes:
   Isobaric: \( W = -P \Delta V \).
   Isochoric: \( W = 0 \).
  Isothermal: \( W = -P V_1 \ln\left(\frac{V_2}{V_1}\right) \).

Answer 1

Correct Answer: 620

The process involves three steps as indicated:
   \(1 \rightarrow 2\): Isobaric process
   \(2 \rightarrow 3\): Isochoric process
   \(3 \rightarrow 1\): Isothermal process
The total work done is given by:
\[W = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 1}.\]
For each step:
\[W_{1 \rightarrow 2} = -P(V_2 - V_1),\]
\[W_{2 \rightarrow 3} = 0 \quad (\text{isochoric process}),\]
\[W_{3 \rightarrow 1} = -P_1 V_1 \ln\left(\frac{V_2}{V_1}\right).\]
Substitute the values:
\[W = \left[ -1 \times (40 - 20) + 0 \right] + \left[ -1 \times 20 \ln\left(\frac{20}{40}\right) \right],\]
\[W = -20 + 20 \ln 2.\]
Using \(\ln 2 = 0.3\):
\[W = -20 + 20 \times 2.3 \times 0.3,\]
\[W = -20 + 6.2,\]
\[W = -6.2 \, \text{bar L}.\]
Convert to joules (\(1 \, \text{bar L} = 100 \, \text{J}\)):
\[|W| = 6.2 \, \text{bar L} \times 100 = 620 \, \text{J}.\]