Cp=20.17+0.3665T at temperature 298k to 473k. Calculate dU ,W ,q, dH at constant pressure

To tackle the problem of calculating the changes in internal energy (dU), work done (W), heat transfer (q), and change in enthalpy (dH) at constant pressure, we can start by using the provided heat capacity equation, Cp = 20.17 + 0.3665T, which is valid for the temperature range from 298 K to 473 K. Let's break down the calculations step by step.

Understanding the Relationships

At constant pressure, the following relationships hold:

• dH = dU + PdV
• q = dH (for processes at constant pressure)
• W = -PdV (work done by the system)

Calculating dH

To find dH, we first need to calculate the change in heat capacity (Cp) over the temperature range. The change in enthalpy can be calculated using the integral of Cp with respect to temperature:

dH = ∫(Cp) dT

Substituting the expression for Cp:

dH = ∫(20.17 + 0.3665T) dT

Now, we can evaluate this integral from T1 = 298 K to T2 = 473 K:

dH = [20.17T + 0.18325T²] from 298 to 473

Calculating the values:

• At T = 473 K: dH = 20.17(473) + 0.18325(473)²
• At T = 298 K: dH = 20.17(298) + 0.18325(298)²

Evaluating the Integral

Now, let's compute these values:

At 473 K:

dH(473) = 20.17 × 473 + 0.18325 × 223729 = 9545.81 + 40919.78 = 50465.59 J

At 298 K:

dH(298) = 20.17 × 298 + 0.18325 × 88804 = 6009.06 + 16250.73 = 22259.79 J

Now, the change in enthalpy (dH) is:

dH = dH(473) - dH(298) = 50465.59 - 22259.79 = 28205.80 J

Calculating dU

To find the change in internal energy (dU), we can use the relationship between dH and dU at constant pressure:

dU = dH - PdV

Since we are considering a constant pressure process, we need to know the volume change (dV). However, if we assume an ideal gas behavior and that the volume change is negligible, we can approximate:

dU ≈ dH

Thus, we can say:

dU ≈ 28205.80 J

Finding Work Done (W)

For work done at constant pressure, we can use the formula:

W = -PdV

Without specific volume change data, we can't calculate W directly. If we assume that the process is isochoric (constant volume), then W = 0. If we had specific volume data, we could calculate it accordingly.

Calculating Heat Transfer (q)

Since we established that at constant pressure, q = dH, we have:

q = 28205.80 J

Summary of Results

In summary, based on the calculations:

• dH = 28205.80 J
• dU ≈ 28205.80 J
• W = 0 J (assuming no volume change)
• q = 28205.80 J

This analysis provides a comprehensive understanding of the thermodynamic changes occurring in the system as it transitions from 298 K to 473 K at constant pressure. If you have any further questions or need clarification on any part of the process, feel free to ask!