Internal Energy and the First Law of Thermodynamics Problems and Solutions 2

  Q#19.11

The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mole of an ideal gas. 

(a) What was the lowest temperature the gas reached in this process? Where did it occur? 

(b) How much work was done by or on the gas from a to b? From b to c? 

(c) If 215 J of heat was put into the gas during abc, how many of those joules went into internal energy?

Answer:
Part ab is isochoric, but bc is not any of the familiar processes. 

pV = nRT determines the Kelvin temperature of the gas. 

The work done in the process is the area under the curve in the pV diagram. Q is positive since heat goes into the gas. 

1 atm = 1.013 $\times 10^5$ Pa; 1 L = $10^{-3} \ m^3$

(a) The lowest T occurs when pV has its smallest value. This is at point a, and

$T_a=\frac{p_aV_a}{nR}$ 

$T_a=\frac{(0.20 \ atm)(1.013 \times 10^5 \ Pa/atm)(2.0 L)(10^{-3} \ m^3/L)}{(0.0175 \ mole)(8.315 J/mol.K)}$

$T_a$ = 278 K

(b) a to b: ΔV = 0 so W = 0. 

b to c: The work done by the gas is positive since the volume increases. 

The magnitude of the work is the area under the curve so 

W = $\frac{1}{2}$(0.50 atm - 0.30 atm)(6.0 L 0 - 2.0 L) = 1.6 L.atm and  

W = (1.6 L.atm)($10^{-3} \ m^3/L$)(1.013 $\times 10^5$ Pa/atm)

= 162 J

(c) For abc, W = 162 J. 

ΔU = Q - W = 215 J - 162 J = 53 J

215 J of heat energy went into the gas. 53 J of energy stayed in the gas as increased internal energy and 162 J left the gas as work done by the gas on its surroundings. 

Q#19.12

A gas in a cylinder is held at a constant pressure of 1.80 $\times 10^5$ Pa and is cooled and compressed from 1.70 $m^3$ to 1.20 $m^3$. The internal energy of the gas decreases by 1.40 $\times 10^5$ J.

 (a) Find the work done by the gas. 

(b) Find the absolute value $|Q|$ of the heat flow into or out of the gas, and state the direction of the heat flow. 

(c) Does it matter whether the gas is ideal? Why or why not? 

Answer:

Calculate W using the equation for a constant pressure process. Then use ΔU = Q − W to calculate Q. 

Given: constant pressure of p = 1.80 $\times 10^5$ Pa

$V_1$ = 1.70 $m^3$ and $V_2$ = 1.20 $m^3$

The internal energy of the gas decreases by $\Delta U$ = 1.40 $\times 10^5$ J

(a) We use W = $\int_{V_1}^{V_2}pdV=p(V_2-V_1)$ for this constant pressure process.

W = (1.80 $\times 10^5$ Pa)(1.20 $m^3-$ 1.70 $m^3$)

W = $-9.00 \times 10^4 \ J$

 (The volume decreases in the process, so W is negative.)

(b)  ΔU = Q − W so, 

Q = ΔU + W = $-1.4 \times 10^5$ J + $-9.00 \times 10^4 \ J$ 

Q = $-2.30 \times 10^5 \ J$

Negative Q means heat flows out of the gas.

(c) W = $\int_{V_1}^{V_2}pdV=p(V_2-V_1)$  (constant pressure) and ΔU = Q − W apply to any system, not just to an ideal gas. We did not use the ideal gas equation, either directly or indirectly, in any of the calculations, so the results are the same whether the gas is ideal or not.   

Q#19.13

BIO Doughnuts: Breakfast of Champions! A typical doughnut contains 2.0 g of protein, 17.0 g of carbohydrates, and 7.0 g of fat. The average food energy values of these substances are 4.0 kcal/g for protein and 9.0 kcal/g carbohydrates and for fat. 

(a) During heavy exercise, an average person uses energy at a rate of 510 kcal/h. How long would you have to exercise to “work off ” one doughnut? 

(b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be 60 kg, and express your answer in and in  km/h.

Answer:

Calculate the total food energy value for one doughnut. K = $\frac{1}{2}mv^2$, 1 cal = 4.186 J

(a) The energy is

(2.0 g)(4.0 kcal/g) (17.0 g)(4.0 kcal/g) (7.0 g)(9.0 kcal/g) = 139 kcal 

The time required is (139 kcal)/(510 kcal/h) = 0.273 h = 16.4

(b) K = $\frac{1}{2}mv^2$, so

v = $\sqrt{\frac{2K}{m}}$

v = $\sqrt{\frac{2(139 \times 10^3 \ cal)(4.186 \ J/cal)}{60 \ kg}}$

v = 139 m/s = 501 km/h

When we set K = Q, we must express Q in J, so we can solve for v in m/s.  

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