Elementos de electromagnetismo
Páginas: 2 (335 palabras)
Publicado: 24 de junio de 2010
19.13: a) . (b (heat flows out of the gas). c) No; the first law of thermodynamics is valid for any system.
19.20: a)
19.23:
a) For constant p,
soheat goes into gas.
b) For constant V,
Q > 0 so heat goes into gas.
c) For constant p,
Q > 0 so heat goes into gas.
19.37: a)
b) From Eq. (19.25),19.39: See Exercise 19.32. a)
b) Using Equation (19.26),
The final temperature is lower than the initial temperature, and the gas is cooled.
17.55: The mass ofwater that the camel saves is
which is a volume of 3.45 L.
17.57: The algebra reduces to
20.5: a)
20.21:
20.39: The initial volume is a)At point 1, the pressure
is given as atmospheric, and with the volume found above,
20.45: There are many equivalent ways of finding the efficiency; the method presented heresaves some steps. The temperature at point 3 is
where has been used for an ideal gas. The work done by the gas during one cycle is the area enclosed by the blue square in Fig.(20.22), and so the efficiency is
20.48: For a monatomic ideal gas,
a) ab: The temperature changes by the same factor as the volume, and so
The work is the same exceptfor the factor of
bc: The temperature now changes in proportion to the pressure change, and and the work is zero
ca: The easiest way to do this is to find the work donefirst; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is and so thework is for the cycle, anticipating part (b)), and so Q must be
b) See above;
c) The heat added, during process ab and ca, is 2.25 and the efficiency is
19.20: a)
19.23:
a) For constant p,
soheat goes into gas.
b) For constant V,
Q > 0 so heat goes into gas.
c) For constant p,
Q > 0 so heat goes into gas.
19.37: a)
b) From Eq. (19.25),19.39: See Exercise 19.32. a)
b) Using Equation (19.26),
The final temperature is lower than the initial temperature, and the gas is cooled.
17.55: The mass ofwater that the camel saves is
which is a volume of 3.45 L.
17.57: The algebra reduces to
20.5: a)
20.21:
20.39: The initial volume is a)At point 1, the pressure
is given as atmospheric, and with the volume found above,
20.45: There are many equivalent ways of finding the efficiency; the method presented heresaves some steps. The temperature at point 3 is
where has been used for an ideal gas. The work done by the gas during one cycle is the area enclosed by the blue square in Fig.(20.22), and so the efficiency is
20.48: For a monatomic ideal gas,
a) ab: The temperature changes by the same factor as the volume, and so
The work is the same exceptfor the factor of
bc: The temperature now changes in proportion to the pressure change, and and the work is zero
ca: The easiest way to do this is to find the work donefirst; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is and so thework is for the cycle, anticipating part (b)), and so Q must be
b) See above;
c) The heat added, during process ab and ca, is 2.25 and the efficiency is
Leer documento completo
Regístrate para leer el documento completo.