Question

ex: calculate the entropy change when 5 moles h₂
underwent an isometric process from 25°c to 100°c. please
think how to calculate the δu and δh, q and w.

Explanation:

Step1: Recall Isometric (Constant Volume) Process

For an isometric process, the volume \( V \) is constant, so work done \( W = 0 \) (since \( W = P\Delta V \) and \( \Delta V = 0 \)).

Step2: Calculate \( \Delta U \) (Internal Energy Change)

For a diatomic gas like \( H_2 \), the molar heat capacity at constant volume \( C_V=\frac{5}{2}R \) (where \( R = 8.314\ J/(mol\cdot K) \)). The formula for \( \Delta U \) is \( \Delta U = nC_V\Delta T \).
First, convert temperatures to Kelvin: \( T_1 = 25^\circ C + 273.15 = 298.15\ K \), \( T_2 = 100^\circ C + 273.15 = 373.15\ K \).
\( \Delta T = T_2 - T_1 = 373.15 - 298.15 = 75\ K \).
\( n = 5\ mol \), so \( \Delta U = 5\ mol\times\frac{5}{2}\times8.314\ J/(mol\cdot K)\times75\ K \).
Calculate: \( \frac{5}{2}\times8.314 = 20.785 \), then \( 5\times20.785\times75 = 5\times1558.875 = 7794.375\ J = 7.794\ kJ \).

Step3: Calculate \( Q \) (Heat Transfer)

In isometric process, \( Q_V = \Delta U \) (from first law \( \Delta U = Q + W \), and \( W = 0 \)), so \( Q = 7.794\ kJ \).

Step4: Calculate \( \Delta H \) (Enthalpy Change)

For a diatomic gas, \( C_P = C_V + R=\frac{5}{2}R + R=\frac{7}{2}R \). The formula for \( \Delta H = nC_P\Delta T \).
\( C_P=\frac{7}{2}\times8.314 = 29.099\ J/(mol\cdot K) \).
\( \Delta H = 5\ mol\times29.099\ J/(mol\cdot K)\times75\ K = 5\times2182.425 = 10912.125\ J = 10.912\ kJ \).

Step5: Calculate Entropy Change (\( \Delta S \))

For isometric process, \( \Delta S = nC_V\ln\frac{T_2}{T_1} \).
\( C_V=\frac{5}{2}\times8.314 = 20.785\ J/(mol\cdot K) \).
\( \ln\frac{373.15}{298.15}=\ln(1.2516)\approx0.224 \).
\( \Delta S = 5\ mol\times20.785\ J/(mol\cdot K)\times0.224 \approx 5\times4.656 = 23.28\ J/K \).

Answer:

• \( \Delta U = \boldsymbol{7.79\ kJ} \) (or \( 7794\ J \))
• \( Q = \boldsymbol{7.79\ kJ} \) (or \( 7794\ J \))
• \( \Delta H = \boldsymbol{10.91\ kJ} \) (or \( 10912\ J \))
• Entropy Change \( \Delta S = \boldsymbol{23.3\ J/K} \) (approximate)