Question

1 mole of Argon gas is heated at constant pressure from 200 K to 600 K.
If \( C_p = 4\ \text{cal} \cdot \text{deg}^{-1} \cdot \text{mol}^{-1} \), then the change in entropy will be:
(Given \( \ln 3 = 1.09 \))

• A. 4.36 Cal.deg$^{-1}$mol$^{-1}$
• B. 2.76 Cal.deg$^{-1}$mol$^{-1}$
• C. 3.34 Cal.deg$^{-1}$mol$^{-1}$
• D. 5.39 Cal.deg$^{-1}$mol$^{-1}$

Hint:

When heating a gas at constant pressure, use $\Delta S = n C_p \ln(T_2/T_1)$. Don’t forget to convert temperature to the same units (Kelvin) before using the formula.

Answer 1

The Correct Option is A

To calculate the change in entropy ($\Delta S$) when heating a gas at constant pressure, we use the thermodynamic relation:
\[ \Delta S = n C_p \ln\left( \frac{T_2}{T_1} \right) \]
Where:
$n$ = number of moles = 1
$C_p$ = specific heat at constant pressure = $4\ \text{Cal.deg}^{-1}\text{mol}^{-1}$
$T_1 = 200\ \text{K}$, $T_2 = 600\ \text{K}$
Calculate the temperature ratio:
\[ \frac{T_2}{T_1} = \frac{600}{200} = 3 \]
Substitute values into the entropy equation:
\[ \Delta S = 1 \cdot 4 \cdot \ln(3) = 4 \cdot 1.09 = 4.36\ \text{Cal.deg}^{-1}\text{mol}^{-1} \]
Thus, the entropy change for the given process is:
\[ \boxed{4.36\ \text{Cal.deg}^{-1}\text{mol}^{-1}} \]