Calculating Internal Energy Change in Ideal Gas Expansion

This solution details the calculation for the change in internal energy ($\Delta U$) of an ideal gas undergoing an expansion process, following the relation $PV^{1.3} = \text{constant}$.

Given Information

• Internal Energy Equation: $U = 3.5 PV + k$
• Initial Volume: $V_1 = 0.25 \, m^3$
• Final Volume: $V_2 = 0.86 \, m^3$
• Initial Pressure: $P_1 = 5 \, N m^{-2}$
• Process Relation: $PV^{1.3} = \text{constant}$
• Constant $k$

Step-by-step Solution

1. Calculate Initial Internal Energy ($U_1$):

   Substitute the initial values ($P_1, V_1$) into the internal energy equation:

   $U_1 = 3.5 P_1 V_1 + k$

   $U_1 = 3.5 \times (5 \, N m^{-2}) \times (0.25 \, m^3) + k$

   $U_1 = 3.5 \times 1.25 \, J + k$

   $U_1 = 4.375 \, J + k$

2. Calculate Final Pressure ($P_2$):

   Using the process relation $P_1 V_1^{1.3} = P_2 V_2^{1.3}$, we find $P_2$:

   $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{1.3}$

   $P_2 = 5 \, N m^{-2} \left( \frac{0.25 \, m^3}{0.86 \, m^3} \right)^{1.3}$

   $P_2 = 5 \times (0.290697...)^{1.3}$

   $P_2 \approx 5 \times 0.19898 \, N m^{-2}$

   $P_2 \approx 0.9949 \, N m^{-2}$

3. Calculate Final Internal Energy ($U_2$):

   Substitute the final volume ($V_2$) and the calculated final pressure ($P_2$) into the internal energy equation:

   $U_2 = 3.5 P_2 V_2 + k$

   $U_2 = 3.5 \times (0.9949 \, N m^{-2}) \times (0.86 \, m^3) + k$

   $U_2 \approx 3.5 \times 0.8556 \, J + k$

   $U_2 \approx 2.9946 \, J + k$

4. Calculate the Change in Internal Energy ($\Delta U$):

   Find the difference between the final energy ($U_2$) and the initial energy ($U_1$):

   $\Delta U = U_2 - U_1$

   $\Delta U = (2.9946 \, J + k) - (4.375 \, J + k)$

   $\Delta U \approx 2.9946 \, J - 4.375 \, J$

   $\Delta U \approx -1.380 \, J$

The calculated change in internal energy is approximately -1.380 Joules.