when 2 mole of an ideal gas(c_pm=R5/2 )heated from 300k to 600k at a const. P. The change in entropy is ?

To find the change in entropy when 2 moles of an ideal gas are heated at constant pressure, we can use the formula for entropy change in a process involving an ideal gas. The formula for the change in entropy (ΔS) at constant pressure is given by:

Entropy Change Formula

ΔS = n * Cp * ln(T2/T1)

Where:

• ΔS = change in entropy
• n = number of moles of the gas
• Cp = molar heat capacity at constant pressure
• T1 = initial temperature
• T2 = final temperature
• ln = natural logarithm

Given Values

In this case:

• n = 2 moles
• T1 = 300 K
• T2 = 600 K
• Cp = R * (5/2) = (5/2) * 8.314 J/(mol·K) = 20.785 J/(mol·K)

Calculating the Change in Entropy

Now, we can plug in the values into the entropy change formula:

ΔS = n * Cp * ln(T2/T1)

ΔS = 2 * 20.785 J/(mol·K) * ln(600 K / 300 K)

First, calculate the natural logarithm:

ln(600/300) = ln(2) ≈ 0.693

Now substitute this back into the equation:

ΔS = 2 * 20.785 J/(mol·K) * 0.693

ΔS ≈ 28.866 J/K

Final Result

The change in entropy for the process of heating 2 moles of the ideal gas from 300 K to 600 K at constant pressure is approximately 28.87 J/K.

This result indicates that as the gas is heated, the disorder or randomness of the gas molecules increases, which is reflected in the positive change in entropy. This aligns with the second law of thermodynamics, which states that the entropy of an isolated system always tends to increase over time.