Question

An ideal gas described by Ti = 300 K, Pi = 2.00 bar, and Vi = 15.0 L is heated at constant volume until P2 = 20.0 bar. It then undergoes a reversible isothermal expansion until P3 = 2.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed-cycle process in a P-V diagram. Calculate number of moles n ( two decimals, as 3.21) = [n]; V2 (unit L, no decimal, such as 11)= [V2]; T2 (unit K, no decimal) = [T2], V3 (unit L, no decimal)= [V3], and T3 (unit K, no decimal) =[T3].

but mostly this one below. Continue with the last question, calculate work (unit J) for step 1, 2 and 3, respectivly: W1= [W1]; W2= [W2]; W3= [W3] . Please enter your answers with 2 decimals in E notation, such as 2.33E4 (refer to 23300). If the answer is negative, please do not forget the negative sign. If answer is zero, please just enter 0 without decimal.

Answer 1

For ideal gas Ti = 300 K, Pi = 2.00 bar, and Vi = 15.0 L

The number of moles n = PiVi/RTi

Where R = gas constant = 0.082 L atm K−1 mol−1

n = (2 x 15)/(0.082 x 300) = 1.22 mol

For step1:

W = 0, because volume is constant

For step 2,

Before calculating the work in step 2, we first calculate T₂.

T₂ = T_i (P₂/P_i)

    = 300 (20.0/2) = 3000K

W = -nRT₂ ln (P₂/P_i)

   = -1.22 mol x 8.314 J mol K x 3000K ln (20/2)

   = -1.22 mol x 8.314 J mol K x 3000K x 2.3 =- 69987.25J =- 6.99 E4J

For step 3,

Calculate V3

P3V₃ = P₂V₂

V3 = 20 bar x 15L /2 bar = 150L

[W3] = -P x Δ V =

          = -2.0 x10^5 Pascal x -1.35 x 10 ^-1 m3/L

          = 2.70 E4 J

W= [W1] + [W2] + [W3]

W = 0 + - 6.99 E4 J + 2.70 E4 J = - 4.99 E4 J