Question

A) Find the pressure of the He at point a.

B) Find the temperature of the He at points a, b, and c.

C) How much heat entered or left the He during segments ab, bc, and ca?

D) By how much did the internal energy of the He change from a to b, from b to c, and from c to a?

Answer 1

All equation numbers assume you have the 14th edition of the textbook.

Part A: Find the pressure of the Helium at point a

Use the ideal gas law PV = nRT where R is the ideal gas constant. The key is that you know part ca is isothermal, which means the temperature T is the same at points a and c. So, find the temperature at C first.

P_cV_c=nRT       --algebra-->       P_c*V_c / (n*R) = T_a = T_c

(2*10⁵_Pa * 0.04_m³) / (3.25_mol * 8.31446_m^3_Pa_°K^-1_mol^-1) = 296_°K

Then, since you now know V, T, and n, use P=nRT/V_a to find P_a. (Hint: V_a=0.01_m^3 is given on the graph). I got around 8*10^5 Pascals.

Part B: Find the temperature of the Helium at points a, b, and c.

Well, you already know T_a = T_c = 296_°K from the previous section. Now just find T_b using PV=nRT. I get T_b = P_bV_b/nR = 1184_°K

Part C: How much heat entered or left the Helium during segments ab, bc, and ca?

For Q_ab, use Q = nCΔT (equation 17.18) where C is the molar heat capacity of Helium. For C, use C=C_p=20.78_J_mol^-1_°K^-1 as given in table 19.1, since this is a constant pressure process on a monatomic gas. This quantity is positive as external heat enters the gas. I get around 6*10^4_J

For Q_bc, do the same except use C_v=12.47_J_mol^-1_°K^-1 since the volume doesn't change along this part. This quantity is negative as heat leaves the gas. I get around -3.6*10^4_J

For Q_ca, it's a little more painful. Part ca is an isothermal process, which means the temperature does not change. But the temperature is being kept constant by a certain amount of heat entering/leaving the system to keep that temperature constant. How on earth do we find Q then? But the textbook gives an important clue in section 19.5: Since this is an ideal gas, if temperature is constant, then internal energy doesn't change. 

ΔU = 0, and since ΔU = Q - W (first law of thermodynamics), then Q = W

So, to find Q, find W. And to find W, you need to integrate the area under the curve from point c to a. The equation of this curve can be found using the ideal gas law, PV=nRT. --algebra-->

P=nRT/V

So, do the indefinite integral of P from V_i to V_f, or 

Q_ca = n*R*T*Integral(1/V, V_i, V_f)

Q_ca = 3.25_mol * 8.31446_m^3_Pa_°K^-1_mol^-1 * 296_°K * evaluate[log(x), from x=0.040 to x=0.010] = -1.11*10⁴_J

Part D: By how much did the internal energy of the Helium change from a to b, from b to c, and from c to a?

Remember:  ΔU = Q - W where U is the internal energy of the system, Q is the heat entering, and W is the work done by the system on its surroundings (volume increase). See section 19.5 for some helpful hints for this section.

ΔU_ab=Q_ab-W_ab. Find the work done. Since this is an isobaric (constant pressure) process, finding the area under the curve is easy: W_ab=P_a*(V_c-V_a). Then, use the value of Q_ab from the previous section to find U

ΔU_bc=Q_bc-W_bc. Since this is an isochoric (constant volume) process, W_bc=0. Then, use the value of Q_bc from the previous section to find U

ΔU_ca=Q_bc-W_bc. Since this is an isothermal (constant temperature) process on an ideal gas, internal energy does not change as explained in the textbook. U is just zero :)

Answer 2

A) p1v1=p2v2
2* 0.04= p2* 0.01
Therefore, p2= 8,
B) Va/Ta= Vb/Tb,
Ta= Pa*Va/n*R = 8*0.01/(3.25*8.314) = 296.07K
Tb= 0.04* 296.07/0.01 = 1184.28,
Tc = 2* 1184.28/8 = 296.07
c) heat in ab= 8 *10^5* (.03)= 24000,
heat in ca = nrT ln (Va/Vc)
D) Internal energy in ca= 0,