15-3. The work done to compress one mole of a monatomic ideal gas is 6200 J. The temperature of the gas changes from 350 to 550 K. (a) How much heat flows between the gas and its surroundings? (b) Determine whether the heat flows into or out of the gas.

(a) An ideal gas has the special property that its internal energy depends only on its temperature. The internal energy of one mole of a monatomic ideal gas at temperature T is

U = (2/3)RT

The change in the internal energy of the gas, as its temperature changes from 350 to 550 K is

DU = (3/2)R(550 - 350) = (3/2)(8.314 J/mole-K)(200 K) = 2490 J

This is related to the heat absorbed by the gas and the work done by the gas through the first law

DU = Q - W

In this case, if 6200 J is the work done on the gas

2490 J = Q - (-6200 J)

Q = -3700 J

(b) Since this value is negative, the direction of heat flow is out of the gas.

15-14. The pressure and volume of a gas are changed along the path ABCA (see page 452 of Cutnell and Johnson). Determine the work done (including the algebraic sign) in each segment of the path: (a) A to B, (b) B to C, and (c) C to A.

In each segment, the work done is PD V, equal to the area under the line on the PV graph. The work done by the gas is positive if the volume of the gas increases.

(a) From A to B, the work done by the gas is (3.0×10⁵ Pa)(3.0 × 10^-3 m²) = 900 J

(b) From B to C, there is no change of volume so the work done is zero.

(c) From C to A, the work done is the area of the trapezoid, which is its width times its average height (or count squares on the graph paper)

W = (1/2)(3.0×10⁵ + 7.0×10⁵ Pa)(-3.0 × 10^-3 m²) = -1500 J

15-24. A monatomic ideal gas (g = 5/3) is compressed adiabatically, and its volume is reduced by a factor of two. Determine the factor by which its pressure increases.

In an adiabatic compression of an ideal gas, the pressure and volume are related by

P_iV_i^g = P_fV_f^g

or

P_f/P_i = (V_i ^g)/(V_f ^g) = (V_i/V_f) ^g

The ratio V_i/V_f is 2, so

P_f/P_i = 2^g = 2^5/3 = 3.175.

15-47. A hiker of mass 58 kg climbs up a mountain through a vertical distance of 950 m. To make the climb, her body generates an extra 4.5 × 10⁶ J of energy. (a) How much work does she do in changing her gravitational potential energy? (b) What is her efficiency (expressed as a fraction between 0 and 1) in doing the work in part (a)?

(a) The increase in her gravitational potential is the same as it was two weeks ago, namely, mgh = (58 kg)(9.8 m/s²)(950 m) = 5.4×10⁵ J.

(b) Her efficiency is the fraction of the total extra energy that her body generates that is converted into useful potential energy. This is

(5.4×10⁵ J)/(4.5 × 10⁶ J) = 0.12

(The rest of the energy is dissipated as heat.)

15-75. Four kilograms of carbon dioxide sublimes from solid "dry ice" to a gas at a pressure of one atmosphere and a temperature of 194.7 K. The latent heat of sublimation is 5.77 × 10⁵ J/kg. Find the change in entropy of the carbon dioxide.

The heat absorbed by the carbon dioxide is

Q = mL = (4 kg)(5.77 × 10⁵ J/kg) = 2.31 × 10⁶ J

The system absorbs this heat at a temperature of 194.7 K, so the change in its entropy is

DS = Q/T = (+2.31 × 10⁶ J)/(194.7 K) = +1.185 × 10⁴ J/K.