Question

One mole of an ideal monoatomic gas is heated in a closed container, first from 273K to 303K, and then from 303K to 373K. The net change in the entropy is ______R. (Rounded off to two decimal places) (R is the ideal gas constant)

Answer 1

Correct Answer: 0.44

Given Data

• Number of moles: \( n = 1 \)
• Ideal gas constant: \( R \)
• For a monoatomic ideal gas, the molar heat capacity at constant volume: \( C_V = \frac{3}{2} R \)
• First temperature change: from \( T_1 = 273K \) to \( T_2 = 303K \)
• Second temperature change: from \( T_2 = 303K \) to \( T_3 = 373K \)

Step 1: Calculate the Entropy Change for the First Process

The change in entropy is given by:

\[ \Delta S = n C_V \ln \frac{T_2}{T_1} \]

Substituting \( n = 1 \), \( C_V = \frac{3}{2} R \), \( T_1 = 273K \), and \( T_2 = 303K \):

\[ \Delta S_1 = \frac{3}{2} R \ln \frac{303}{273} \]

\[ \Delta S_1 \approx \frac{3}{2} R \ln(1.111) \]

\[ \Delta S_1 \approx \frac{3}{2} R \times 0.1054 \]

\[ \Delta S_1 \approx 0.1581 R \]

Step 2: Calculate the Entropy Change for the Second Process

Similarly, for the second temperature change:

\[ \Delta S_2 = \frac{3}{2} R \ln \frac{T_3}{T_2} \]

Substituting \( T_2 = 303K \) and \( T_3 = 373K \):

\[ \Delta S_2 = \frac{3}{2} R \ln \frac{373}{303} \]

\[ \Delta S_2 \approx \frac{3}{2} R \ln(1.231) \]

\[ \Delta S_2 \approx \frac{3}{2} R \times 0.2078 \]

\[ \Delta S_2 \approx 0.3117 R \]

Step 3: Calculate the Total Entropy Change

The total entropy change is the sum of \( \Delta S_1 \) and \( \Delta S_2 \):

\[ \Delta S = \Delta S_1 + \Delta S_2 \]

\[ \Delta S = 0.1581 R + 0.3117 R \]

\[ \Delta S \approx 0.4498 R \]

Final Answer

The net change in entropy is approximately \( 0.44 R \).