Chapter 2, Problem 2.35E

A piston having 7.23 g of steam at 110 °C increases its temperature by 35 °C. At the same time, it expands from a volume of 2.00 L to 8.00 L against a constant external pressure of 0.985 atm. Calculate w, q, ΔU, and ΔH for the process.

Interpretation Introduction

Interpretation:

A piston having 7.23 g of steam at 110 °C increases its temperature by 35 °C. At the same time, it expands from a volume of 2.00 L to 8.00 L against a constant external pressure of 0.985 atm. For the process w, q, ΔU, and ΔH is to be calculated.

Concept introduction:

Larger collection of atoms and molecules can be handled easily with the concepts of thermodynamics. The terms heat, work, internal energy, and enthalpy can be explained well by macroscopic rules. When an object is moving distance s due to the externally applied force F, then it is called some work has been done on an object.

Therefore, work (w) = F. s

In constant external pressure (P_ex), the small amount of work done by the system (dw) to the surroundings with infinitesimal small change of volume (dV) is given by the expression

$\mathrm{d}\mathrm{w}\mathrm{ }=\mathrm{ }-\mathrm{P}\mathrm{e}\mathrm{x}\mathrm{ }\mathrm{d}\mathrm{V}\mathrm{ }$

The decrease of system energy is denoted by the negative sign. Normally, heat is defined as the measurement of thermal energy transfer that can be obtained by the temperature change in an object and is referred by the symbol q. when the heat enters into the system it is getting positive sign and when the heat comes out of the system it obtains negative sign. Notably, the overall energy content of the system is defined as internal energy (ΔU). The internal energy values for an isolated system is zero. since, no heat can enter into the system. Mathematically, internal energy can be written as,

$\mathrm{\Delta }\mathrm{U}\mathrm{ }=\mathrm{ }\mathrm{q}\mathrm{ }+\mathrm{ }\mathrm{w}\mathrm{ }$

Primarily, most of the processes are carried out at constant pressure, instead of constant volume, thus, a new concept of enthalpy was introduced. Thus, the enthalpy (H) of a system, involving pressure volume work, is given by the expression,

H = U + pV

In a given system, the enthalpy and internal energy are governed by state variables of the system.

Answer to Problem 2.35E

For the mentioned process the values of w, q, ΔU, and ΔH is calculated as follows;

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }=\mathrm{ }\mathrm{w}\mathrm{ }=-\mathrm{P}\mathrm{e}\mathrm{x}\mathrm{ }\mathrm{d}\mathrm{V}\mathrm{ }\mathrm{ }=\mathrm{ }-599\mathrm{ }\mathrm{J}$

$\begin{array}{l}\text{Heat of the system = q = 516 J}\hfill \\ \text{Change internal energy =ΔU = -83 J}\hfill \\ \text{Change in enthalpy =ΔH = +34 J}\hfill \end{array}$

Explanation of Solution

The first law thermodynamics is basically including the three fundamental parameters work, heat and internal energy. Enthalpy and internal energy are referred as state functions.

Given,

$\begin{array}{l}\text{Mass of steam = m = 7}\text{.23 g}\hfill \\ \text{Temperature initial = Ti = 110 °C = 383 K}\hfill \\ {\text{Temperature final = Tf}}_{}\text{= 35 °C = 308 K}\hfill \\ \text{Volume initial = Vi = 2 L}\hfill \\ \text{Volume final = Vf = 8 L}\hfill \\ \text{Pressure = Pi = 0}\text{.985 atm}\hfill \end{array}$

From the values of temperature difference of the system, the quantity of heat absorbed by the system q is calculated as = 516 J

Work (w):

$\begin{array}{l}{\text{We know that w = -P}}_{\text{ex}}\text{ΔV, on substituting the values in the expression, we get}\hfill \\ \text{w = - 0}\text{.985 atm }\left(\text{8L – 2L}\right)\hfill \\ \text{ = - 0}\text{.985 atm }\text{. 6 L}\hfill \\ \text{w = - 5}\text{.91 L atm}\hfill \\ \text{1 L atm is equal to 101}\text{.32 J, therefore,}\hfill \\ \text{w = - 599 J}\hfill \\ \text{internal energy (ΔU):}\hfill \\ \text{we know that, first law of thermodynamics is given by the expression,}\hfill \\ \text{q =ΔU + w }\left(\text{or}\right)\hfill \\ \text{ΔU = q – w, on substituting the values in the expression, we get}\hfill \\ \text{ = 516 J – 599 J}\hfill \\ \text{ΔU = - 83 J}\hfill \\ \hfill \\ \hfill \end{array}$

$\begin{array}{l}\text{Enthalpy (ΔH):}\hfill \\ \text{ΔH =ΔU + pΔV, on substituting the values in the expression, we get}\hfill \\ \text{ = -83 J + 117 J}\hfill \\ \text{ΔH = +34 J}\hfill \end{array}$

Conclusion

Thus, For the process w, q, ΔU, and ΔH is calculated.