The First Law of Thermodynamics Problems and Solutions 7

  Q#19.57 

A Thermodynamic Process in an Insect. The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. P19.57). The beetle’s body has reservoirs of two different chemicals; when the beetle is disturbed, these chemicals are combined in a reaction chamber, producing a compound that is warmed from 20$^0C$ to 100$^0C$ by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s (68 km/h) scaring away predators of all kinds. (The beetle shown in the figure is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the two chemicals and the spray is the same as that of water, 4.19 $\times 10^3 \ J/kg.K$, and that the initial temperature of the chemicals is 20$^0C$.

Answer:

The heat produced from the reaction is

$Q_{reaction} = mL_{reaction}$, where $L_{reaction}$ is the heat of reaction of the chemicals.

$Q_{reaction}=W + \Delta U_{spray}$

For a mass m of spray, 

W = $\frac{1}{2}mv^2$ = $\frac{1}{2}m(19 \ m/s)^2=(180.5 \ J/kg)m$ and

$\Delta U_{spray}=Q_{spray} = mc\Delta T=m(4190 \ J/kg.^0C)(100^0C-20^0C)=(335,200 \ J/kg)m$

Then $Q_{reaction}=(180 J/kg + 335,200 \ J/kg)m=(335,380 \ J/kg)m$ and

$Q_{reaction} = mL_{reaction}$  implies

$(335,380 \ J/kg)m = mL_{reaction}$

The mass m divides out and

$L_(reaction)=335,380 \ J/kg \approx 3.4 \times 10^5$ J/kg 

The amount of energy converted to work is negligible for the two significant figures to which the answer should be expressed. Almost all of the energy produced in the reaction goes into heating the compound.

Q#19.58

High-Altitude Research. A large research balloon containing 2.00 $\times 10^3 \ m^3$ of helium gas at 1.00 atm and a temperature of 15.0 $^0C$ rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. P19.58). Assume the helium behaves like an ideal gas and the balloon’s ascent is too rapid to permit much heat exchange with the surrounding air. 

(a) Calculate the volume of the gas at the higher altitude. 

(b) Calculate the temperature of the gas at the higher altitude. 

(c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

Answer:

The process is adiabatic. Apply $p_1V_1^{\gamma}=p_2V_2^{\gamma}$ and pV = nRT. Q = 0, so

$\Delta U = -W=-\frac{1}{\gamma - 1}(p_1V_1-p_2V_2)$

For helium, $\gamma$ = 1.67, $p_1=1.00$ atm = $1.013 \times 10^5$ Pa. $V_1=2.00 \times 10^3 \ m^3$ 

$p_2=0.900$ atm = $9.117 \times 10^4$ Pa. $T_1=288.15 \ K$ 

(a) $V_2^{\gamma}=V_1^{\gamma}\left(\frac{p_1}{p_2}\right)$

$V_2=V_1\left(\frac{p_1}{p_2}\right)^{\frac{1}{\gamma}}$

$V_2=(2.00 \times 10^3 \ m^3)\left(\frac{1.00 \ atm}{0.900 \ atm}\right)^{\frac{1}{1.67}}$

$V_2=2.13 \times 10^3 \ m^3$

(b) pV = nRT gives

$\frac{T_1}{p_1V_1}=\frac{T_2}{p_2V_2}$

$\frac{288.15 \ K}{(1.00 \ atm)(2.00 \ times 10^3 \ m^3)}=\frac{T_2}{(0.900 \ atm)(2.13 \times 10^3 \ m^3)}$

$T_2=276.2 K = 3.0^0C$

(c) $\Delta U =-\frac{1}{1.67 - 1}[(1.013 \times 10^5 \ Pa)(2.00 \times 10^3 \ m^3)-(9.177 \times 10^4 \ Pa)(2.13 \times 10^3 \ m^3)]$

$\Delta U = -1.25 \times 10^7$ J

The internal energy decreases when the temperature decreases.   

Q#19.59 

During certain seasons strong winds called chinooks blow from the west across the eastern slopes of the Rockies and downhill into Denver and nearby areas. Although the mountains are cool, the wind in Denver is very hot; within a few minutes after the chinook wind arrives, the temperature can climb 20$C^0$ (“chinook” is a Native American word meaning “snow eater”). Similar winds occur in the Alps (called foehns) and in southern California (called Santa Anas). 

(a) Explain why the temperature of the chinook wind rises as it descends the slopes. Why is it important that the wind be fast moving? 

(b) Suppose a strong wind is blowing toward Denver (elevation 1630 m) from Grays Peak (80 km west of Denver, at an elevation of 4350 m), where the air pressure is 5.60 $\times 10^4$ Pa and the air temperature is  $-15.0^0C$. The temperature and pressure in Denver before the wind arrives are 2.0$^0C$ and 8.12 $\times 10^4$ Pa. By how many Celsius degrees will the temperature in Denver rise when the chinook arrives?

Answer:

For an adiabatic process of an ideal gas,

$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ and pV = nRT. Q = 0, so

$\Delta U = -W=-\frac{1}{\gamma - 1}(p_1V_1-p_2V_2)$

For air, $\gamma$ = 1.40 = 7/5

(a) As the air moves to lower altitude its density increases; under an adiabatic compression, the temperature rises. If the wind is fast-moving, Q is not as likely to be significant, and modeling the process as adiabatic (no heat loss to the surroundings) is more accurate.

(b) $V = \frac{nRT}{p}$, so

$T_1V_1^{\gamma-1}=T_2V_2^{\gamma}$ gives 

$T_1^{\gamma}p_1^{1-\gamma}=T_2^{\gamma}p_2^{1-\gamma}$

The temperature at the higher pressure is

$T_2=T_1\left(\frac{p_1}{p_2}\right)^{(\gamma-1)/\gamma}$

$T_2=(258.15 \ K)\left(\frac{8.12 \times 10^4 \ Pa}{5.60 \times 10^4 \ Pa}\right)^{2/7}$

$T_2=287.1 \ K = 13.9^0C$

 so the temperature would rise by $11.9^0C$

In an adiabatic compression, Q = 0 but the temperature rises because of the work done on the gas. 

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