Answer:
Explanation:
check the image
a)
The workdone on the gas during during the isobaric compression
W = pΔV
= nRΔT
= 2 * 8.314 * (273-110)
=2710J
b) The workdone on the gas during during the isothermal expansion
[tex]W= nRT_fIn(\frac{V_f}{V_i} )\\\\2*8.314*110In\frac{110}{273} \\\\=-1662.6J[/tex]
Answer:
a) The work done during the isobaric compression is 4060.91 J.
b) The work done during the isothermal expansion is 2489.58 J.
Explanation:
a) Here we have two different processes. We calculate the work done during the isobaric compression using the following formula:
[tex] W= -P. (V₂ - V₁) [/tex]
Since the process is isobaric, the pressure is constant. First we need to find this value. In order to do that, we use the ideal gas equation:
[tex] p.V= n.R.T [/tex]
[tex] p= [n.R.T] ÷ V = [3 mol × 0.082 (L.atm÷mol.K) × 273 K] ÷ 3 L [/tex]
p = 22.39 atm
Now, we need to find V₂. We do this using Charles' law:
[tex] V₁/T₁ = V₂/T₂ [/tex]
[tex] V₂ = V₁/T₁ × T₂ = 3 L ÷ 273 K × 110 K [/tex]
V₂ = 1.21 L
Now we are in condition to calculate the work done during the isobaric compression:
[tex] W= -P. (V₂ - V₁) [/tex]
[tex] W= -22.39 atm. (1.21 - 3) L [/tex]
W= 40. 07 L.atm
We multiply this value by 101.325 to convert it to joules: W= 4060.91 J
b) To calculate the work done during the isothermal expansion, we use the following equation:
[tex] W = n.R.T. ln (V₂/V₁) [/tex]
[tex] W = 3 mol ×0.082 (L.atm/mol.K) × 110 K × ln (3L / 1.21 L) [/tex]
W= 24.57 L.atm
We convert this value to joules: W= 2489.58 J
