EJERCICIOS DE TERMODINAMICA

Problemas Capítulo 1.

1. Calculate the work performed by a body expanding from an initial volume of 3.12 liters to a final volume of 4.01 liters at the pressure of 2.34 atmospheres.

Se sustituyen los datos en la integral para un cambio finito de volumen de V1 a V2

W= ʃ pdV

Se obtiene el trabajo efectuado en un cambio de volumen a presión constante.

W= p (V2-V1)

W= 2.34 atm (4.01 lt- 3.12 lt)

W= (237100.5 Kg/ms²) (0.00401m³-.00312m³)

W= 211.01 Kg/mS². m³

W=211.01 Kgm²/ s² = Energía (la p y el V son 2 variables canónicamente

conjugadas)

2. Calculate the pressure of 30 grams of hydrogen inside a container of 1 cubic meter at the temperature of 18°C

Sustituimos los valores en la ecuación de los gases ideales teniendo la constante de los gases R= 8314472 J/mol.K

pV= m/M (RT)

p=m RT/V

p= 30 moles/ 1m³ (8314472 J/mol.K x 291 °K)

p= 7.25 x 10¹º J/m³

p= 7.25 x10¹º Kg/ ms²

3. Calculate the density and specific volume of nitrogen at the temperature of 0°C

4. Calculate the work performed by 10 grams of oxygen expanding isothermally at 20°C from 1 to .3 atmospheres of pressure

De la ecuación del trabajo efectuado en un cambio de volumen sustituimos la presión de n moles de gas ideal.

W= ʃ pdV donde p= nRT

Sustituimos esto en la integral, y sacamos las constantes n, R y T y calculamos la integral:

W= n RT ʃ dV/V = nRT ln V2/V1

Cuando T es constant p1V1= p2V2 o bien V2/V1=p/P2

Asi que el trabajo isotérmico también puede expresarse como:

W= nRT ln p2/p1 (gas ideal, proceso isotérmico)

W= 10 mol (8314472 J/mol.K) (273 °K) Ln(303975 Kg/ms²/101325 Kg/ms²)

W= 1.77x10¹º Ln 3

W=1.9293x10¹º Kgm²/ s² = energia

Problemas capitulo 2.

1. Calculate the energy variation of a sistem which performs 3.4 X 10⁸ ergs of work and absorbs 32 calories of heat.

2. How many calories are absorbed by 3 moles of an ideal gas expanding isothermally from the initial pressure of 5 atmospheres to the final pressure of 3 atmospheres, at the temperature of 0°C?

3. One mole of a diatomic ideal gas performs a transformation from an initial state for which temperature and volume are, respectively, 291°K and 21000 cc. to a final state in which temperature and volume are 305°K and 12700 cc. The transformation is represented on the (V,p) diagram by a straight line. To find the work performed and the heat absorbed by the system.

4. 4. A diatomic gas expands adiabatically to a volume 1.35 times larger than the initial volume. The initial temperature is 18°C. Find the final temperature.

Problemas Capitulo 3.

1. One mole of a monatomic gas performs a Carnot cycle between the temperaturas 400°K and 300°K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume 5 liters. To find the work performed during a cycle, and the amounts of heat exchanged with the two sources.

2. What is the maximum efficiency of a thermal engine working between an upper temperature of 400°C and a lower temperature of 18°C?

3. Find the minimum amount of work needed to extract one calorie of heat from a body at the temperature of 0°F, when the temperature of the environment is 100°F.

Problemas Capitulo 4.

1. What is the entropy variation of 1000 grams of water when raised from freezing to boiling temperature? (assume a constant specific heat= 1 cal/gm.deg)

2. A body obeys the equation of state:

pV¹˙²= 10⁹T¹˙¹

A measurement of its thermal capacity inside a container having the constant volume 100 liters shows that under these conditions, the thermal capacity is constant and equal to 0.1 cal/deg. Express the energy and the entropy of the system as functions of T and V.

3. The boiling point of ethyl alcohol (C₂H₆O) is 78.3°C; the heat of vaporization is 855 Joules/gm. Find dp/dT at the boiling point.

Problemas capitulo 5.

1. With the aid of the phase rule discuss the equilibrium of a saturated solution and the solid of the dissolved substance.

2. How many degrees of freedom has the system composed of a certain amount of water and a certain amount of air? (Neglect the rare gases and the carbon dioxide contained in air)

3. The electromotive force of a reversible electric cell, as a function of the temperature, is:

0.924 + 0.0015t + 0.0000061t² voltst being the temperature in °C. Find the heat absorbed by the cell when one coulomb of electricity flows through it isothermally at a temperature of 18°C.