The graph in the figure (Figure 1) shows a p V diagram for 3.30moles of ideal he

Question

The graph in the figure (Figure 1) shows a pV diagram for 3.30moles of ideal helium (He) gas. Part ca of this process is isothermal.

C)How much heat entered or left the He during segments ab, bc, and ca?Express your answers using three significant figures separated by commas

D)In each segment, did the heat enter or leave?

E)By how much did the internal energy of the Hechange from a to b, from b to c, and from c toa? Express your answers using three significant figures separated by commas

F)Indicate whether this energy increased or decreased.

Explanation / Answer

a) As ca is an isothermal process, T is same for both c and a, For point c, Pc = 2 x 10^5 Pa. Vc = 0.04cu. m. Then. PV = nRT, T = (2 x 10^5 x 0.04)/(3.3 x 8.31) = 291.7 K.

At A, Pa = ?, Va = 0.01 cu. m. Ta = 291.7 K. Then, Pa = nRTa/Va = 3.3 x 8.31 x 291.7/0.01 = 7.986 x 10^5 Pa

b) Ta = 291.7 K, Tc = 291.7 K, Tb = Pb x Vb/nR.

Pa = Pb = 7.986 x 10^5 Pa

Then, Tb = 7.986 x 10^5 x 0.04 / (3.3 x 8.31) = 1164.8 K

c) ab -> Isobaric process

Qab = nC_p (Tb - Ta) = 3.3 x 5/2 x R x (1164.8 - 291.7) = 6 x 10^4 J.(heat entered)

Uab = nC_v(Tb - Ta) = 3.3 x 3/2 x R x (1164.8 - 291.7) = 3.6 x 10^4 J (energy increased)

bc-> Isochoric process

Qbc = nC_v (Tc - Tb) = 3.3 x 3/2 x 5 x (291.7 - 1164.8) = -2.2 x 10^4 J (heat left the system)

Ubc = Qbc (as work done is zero) = -2.2 x 10^4 J (energy decreased)

ca -> Isothermal process

Qca = nRT ln(Va/Vc) = 3.3 x 8.31 x 1164.8 x ln(0.01/0.04) = -4.4 x 10^4 J (Heat left)

Uca = 0 (no change)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.