One mole of an ideal monoatomic gas undergoes two reversible processes $\left(A\to BandB\to C\right)$ as shown in the given figure:

 $A\to B$ is an adiabatic process. If  the total heat absorbed in the entire process $\left(A\to BandB\to C\right)$ is $R{T}_2$ ln 10, the value of   $2 \log V_3$  is
[Use, molar heat capacity of the gas at constant pressure, $C_{p,m}$ = $\frac{5}{2}$ R]

Thermodynamics Problem: Ideal Monoatomic Gas

Given:

• One mole of ideal monoatomic gas
• A → B: Adiabatic process
• B → C: Isothermal process at T₂
• C_P,m = (5/2)R
• Q_total = RT₂ ln 10
• From diagram: T₁ = 600 K, V₁ = 10 m³, T₂ = 60 K

Step 1: Adiabatic Process (A → B)

For an adiabatic process: TV^γ-1 = constant

For monoatomic gas: γ = C_P/C_V = 5/3, so γ - 1 = 2/3

T₁V₁^2/3 = T₂V₂^2/3

V₂/V₁ = (T₁/T₂)^3/2

Step 2: Isothermal Expansion (B → C)

For isothermal process at T₂:

Q_B→C = nRT₂ ln(V₃/V₂)

Since A→B is adiabatic (Q_A→B = 0):

Q_total = Q_B→C = RT₂ ln(V₃/V₂)

Given Q_total = RT₂ ln 10:

RT₂ ln(V₃/V₂) = RT₂ ln 10 ⟹ V₃/V₂ = 10

V₃ = 10V₂

Step 3: Calculate 2 log V₃

From Step 1:

V₂ = V₁(T₁/T₂)^3/2

Therefore:

V₃ = 10V₁(T₁/T₂)^3/2

Taking logarithm:

2 log V₃ = 2 log[10V₁(T₁/T₂)^3/2]

= 2 log 10 + 2 log V₁ + 3 log(T₁/T₂)

Substituting values:

V₁ = 10 ⟹ log V₁ = 1

T₁/T₂ = 600/60 = 10 ⟹ log(T₁/T₂) = 1

2 log V₃ = 2(1) + 2(1) + 3(1) = 2 + 2 + 3 = 7

Final Answer:

The value of 2 log V₃ = 7