For 1 mole of an ideal monatomic gas on moving from one state to another, the temperature is doubled but pressure becomes √2 times. The entropy change in the process will be (R = 2 cal/mol · K)

• (A) R ln 2
• (B) 2 R ln 2
• (C) 3 R ln 2
• (D) R/2 ln 2

To find the entropy change for 1 mole of an ideal monatomic gas when the temperature is doubled and the pressure becomes √2 times, we can use the formula for entropy change:

Entropy Change Formula

The change in entropy (ΔS) can be calculated using the formula:

• ΔS = nR ln(Vf/Vi) + nCvdT/T

Where:

• n = number of moles
• R = universal gas constant
• Cv = molar heat capacity at constant volume
• Vi and Vf = initial and final volumes

Given Conditions

In this case:

• Initial temperature (T1) = T
• Final temperature (T2) = 2T
• Initial pressure (P1) = P
• Final pressure (P2) = √2 P

Finding Volume Change

Using the ideal gas law, we can express the volumes:

• V1 = nRT1/P1 = nRT/P
• V2 = nRT2/P2 = nR(2T)/(√2 P) = (2nRT)/(√2 P) = (√2 nRT)/P

Calculating Entropy Change

Now, substituting the volumes into the entropy change formula:

• ΔS = nR ln(V2/V1) + nCvdT/T
• ΔS = nR ln((√2 nRT)/P / (nRT/P)) + nCvd(2T - T)/T
• ΔS = nR ln(√2) + nCvdT/T

For a monatomic ideal gas, Cv = (3/2)R. Therefore, the second term becomes:

• ΔS = nR ln(√2) + n(3/2)R(2T - T)/T
• ΔS = nR ln(√2) + n(3/2)R

Final Calculation

Substituting n = 1 mole and R = 2 cal/mol·K:

• ΔS = 2 ln(√2) + (3/2)(2)
• ΔS = 2(1/2 ln(2)) + 3
• ΔS = ln(2) + 3

Thus, the entropy change is:

• ΔS = 2R ln(2)

Final Answer

The correct option is (B) 2 R ln 2.