Answer to Question #159926 in General Chemistry for Taylor

Question

A closed system consisting of 1.00 mol of a monoatomic ideal gas has an initial pressure of 1.00 bar and an initial temperature of 25 ̊C. The system undergoes the series of two transitions described below.

1. (8 points) First the system is heated isobarically from 25 ̊C to 125 ̊C. What are the heat, work, and internal energy change for this isobaric heating step? Give your answers in joules.
2. (4 points) Next, the system is compressed isothermally and reversibly back to its original volume. What are the heat, work, and internal energy change for this isothermal reversible compression? Give your answers in joule

Answer 1

Isobaric process

1. For mono-atomic gases, Cp = 5/2 R

Internal energy Δu = nCvΔT .n = 1mol Cv = 3/2R ΔT = 100K

= 1× 3/2 R × 100 = 3/2 × 8 314 ×100

= 1247.1J

Heat = 5/3 × Δu = 5/3 × 1247.1

= 2078.5J

W = Δu - Q

W = 1247.1J - 2078.5J

= -831.4J

2. V1/T1 = V2/T2

= T2/T1 = V2/V1

W_rev = -nRTIn(v2/v1)

W = - 1 × 8.314 × 298 × In(125/25)

W= -3987.5J

Q=-W

Q= 3987.5J

Since the process is isothermal Δu = 0