One mole of an ideal monoatomic gas undergoes two reversible processes (A → B and B → C) as shown in the given figure:

The correct answer is 7.

Step-by-step Explanation:

From the given figure:

We are given the following data from the graph and standard thermodynamic properties:
• Number of moles of ideal monoatomic gas, _n = 1
• Initial state A: Temperature _T1 = 600 K, Volume _V1 = 10 m³
• Intermediate state B: Temperature _T2 = 60 K, Volume = _V2
• Final state C: Temperature = _T2 = 60 K, Volume = _V3
• For a monoatomic ideal gas, the adiabatic index is:

$\gamma =\frac{5}{3}$

Step 1: Process A → B (Reversible Adiabatic Process)

For a reversible adiabatic process, no heat is absorbed, so _QA→B = 0.
Using the adiabatic relationship between temperature and volume:

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

Since $\gamma -1=\frac{5}{3}-1=\frac{2}{3}$, we substitute the given values:

$600\times {\left(10\right)}^{2/3}=60\times V_2^{2/3}$

Simplifying the equation:

$10\times {10}^{2/3}=V_2^{2/3}$

${10}^{5/3}=V_2^{2/3}$

Raising both sides to the power of $\frac{3}{2}$:

$V_2={\left({10}^{5/3}\right)}^{3/2}={10}^{5/2}{\text{m}}^3$

Step 2: Process B → C (Reversible Isothermal Process)

In process B → C, the temperature remains constant at _T2 = 60 K.
The heat absorbed during an isothermal expansion is:

$Q_{B\to C}=nR{T}_2\ln \left(\frac{V_3}{V_2}\right)$

The total heat absorbed in the overall process is _Qtotal = _QA→B + _QB→C = RT₂ ln(10).
Since _QA→B = 0 and _n = 1:

$R{T}_2\ln \left(\frac{V_3}{V_2}\right)=R{T}_2\ln \left(10\right)$

Equating the terms inside the logarithm:

$\frac{V_3}{V_2}=10\Rightarrow V_3=10\times V_2$

Substituting _V2 = 10^5/2:

$V_3=10\times {10}^{5/2}={10}^{7/2}$

Step 3: Calculating 2 log₁₀(V₃)

Taking the logarithm (base 10) of _V3:

${\log }_{10}\left(V_3\right)={\log }_{10}\left({10}^{7/2}\right)=\frac{7}{2}$

Thus, the value of 2 log₁₀(V₃) is:

$2{\log }_{10}\left(V_3\right)=2\times \frac{7}{2}=7$