Chapter 7, Problem 34E

Calculate the internal energy change for each of the following.

a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases 23 J of heat.

b. A piston is compressed from a volume of 8.30 L to 2.80 L against a constant pressure of 1.90 atm. In the process, there is a heat gain by the system of 350. J.

c. A piston expands against 1.00 atm of pressure from 11.2 L to 29.1 L. In the process, 1037 J of heat is absorbed.

Interpretation Introduction

Interpretation: The internal energy change for given cases are calculated.

Concept Introduction:  Internal energy is a total energy of a system, consists of kinetic, potential energies. The internal energy change is accompanied by the heat flow or as work done on or by a system.

Formula:

                                           $\text{ΔE=q}\pm \text{w}\mathrm{......}\text{(1)}$

                                                   q is a heat

                                                    w is a work

Work done by the system means heat and work will be

                                        $\text{ΔE=q-w}\mathrm{......}\text{(2)}$

Work done on the system means heat and work will be

                                        $\text{ΔE=}\text{q+w}\mathrm{......}\text{(3)}$

                                          $\text{w=-PΔV}\mathrm{......}\text{(4)}$

                                                  P is a constant pressure

                                                 $\text{ΔV}$ is a volume change

Answer to Problem 34E

Internal energy   $\text{ΔE=77}\text{J}$

Explanation of Solution

Given data:

                    $\text{Relesed heat q (a) =23J}$

                    Compression work (a) $\text{w=100J}$

                    Volume change $\text{ΔV}$ (b)= $\text{(2}\text{.80+8}\text{.30)}$ $\text{=-5}\text{.50}\text{L}$

                    Constant pressure (b) $\text{=1}\text{.90}\text{atm}$

                    Heat gained q (b) $\text{=350}\text{J}$

                    Volume change $\text{ΔV}$ (c)= $\text{(29}\text{.1-11}\text{.2)}$ $\text{=17}\text{.90 }\text{L}$

                    Constant pressure (c) $\text{=1}\text{.00}\text{atm}$

                    Heat absorbed q (c) $\text{=1037}\text{J}$

The internal energy of (a).

                                                    $\text{ΔE=-23}\text{J+100}\text{J}$

                                                      $\text{ΔE=77}\text{J}$

• The given values are plugging to the equation 4 to get internal energy of (a). The internal energy of (a) is $\text{77}\text{J}$.

Interpretation Introduction

Interpretation: The internal energy change for given cases are calculated.

Concept Introduction:  Internal energy is a total energy of a system, consists of kinetic, potential energies. The internal energy change is accompanied by the heat flow or as work done on or by a system.

Formula:

                                           $\text{ΔE=q}\pm \text{w}\mathrm{......}\text{(1)}$

                                                   q is a heat

                                                    w is a work

Work done by the system means heat and work will be

                                        $\text{ΔE=q-w}\mathrm{......}\text{(2)}$

Work done on the system means heat and work will be

                                        $\text{ΔE=}\text{q+w}\mathrm{......}\text{(3)}$

                                          $\text{w=-PΔV}\mathrm{......}\text{(4)}$

                                                  P is a constant pressure

                                                  $\text{ΔV}$ is a volume change

Answer to Problem 34E

Internal energy   $\Delta \text{E=1408}\text{.6}\text{J}$

Explanation of Solution

Explanation

To calculate the internal energy of (a).

                                                   Work $\text{=-1}\text{.90×(2}\text{.80-8}\text{.30)}$

                                                             $\text{=10}\text{.45}\text{atm}$

                                       For in Joules = $\text{10}\text{.45×101}\text{.3}\text{J}$

                                                              $\text{=1058}\text{.6}\text{J}$

                                  Internal energy $\text{=1058}\text{.6+350}$

                                                             $\text{=1408}\text{.6}\text{J}$

• The given volume change and constant pressure are plugging in to equation (4)to get  work of the piston and this work can added with given heat to get internal energy of (b).
• The internal energy of (b) is $\text{1408}\text{.6}\text{J}$.

Interpretation Introduction

Interpretation: The internal energy change for given cases are calculated.

Concept Introduction:  Internal energy is a total energy of a system, consists of kinetic, potential energies. The internal energy change is accompanied by the heat flow or as work done on or by a system.

Formula:

                                           $\text{ΔE=q}\pm \text{w}\mathrm{......}\text{(1)}$

                                                   q is a heat

                                                    w is a work

Work done by the system means heat and work will be

                                        $\text{ΔE=q-w}\mathrm{......}\text{(2)}$

Work done on the system means heat and work will be

                                        $\text{ΔE=}\text{q+w}\mathrm{......}\text{(3)}$

                                          $\text{w=-PΔV}\mathrm{......}\text{(4)}$

                                                  P is a constant pressure

                                                 $\text{ΔV}$ is a volume change

Answer to Problem 34E

Internal energy  $\Delta \text{E=776}\text{.27}\text{J}$

Explanation of Solution

• The given volume change and constant pressure are plugging in to equation (4)to get  work of the piston and this work can added with given heat to get internal energy of (c).
• The internal energy of (b) is $\text{776}\text{.27}\text{J}$.
• To calculate the internal energy of (c).

•                                              Work $\text{=-1}\text{.00×(29}\text{.1-11}\text{.2)}$

                                                          $\text{=}\text{17}\text{.9}\text{atm}$

                                    For in Joules = $\text{17}\text{.90×101}\text{.3}\text{J}$

                                                          $\text{=1813}\text{.27}\text{J}$

•                               Internal energy $\text{=1813}\text{.27-1037}$ $\text{=}\text{776}\text{.27}\text{J}$