Question

We now consider three moles of ideal gas at the same initial state (3.0 L at 273 K). This time, we will first perform an isobaric compression, then an isothermal expansion to bring the gas to the final state with the same volume of 3.0 L, and at temperature 110 K. Hint (a) How much work (in J) is done on the gas during the isobaric compression? Wcompression = J (b) How much work (in J) is done on the gas during the isothermal expansion? Wexpansion = J

Answer 1

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

$W= nRT_fIn(\frac{V_f}{V_i} )\\\\2*8.314*110In\frac{110}{273} \\\\=-1662.6J$

Answer 2

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:

$W= -P. (V₂ - V₁)$

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:

$p.V= n.R.T$

$p= [n.R.T] ÷ V = [3 mol × 0.082 (L.atm÷mol.K) × 273 K] ÷ 3 L$

p = 22.39 atm

Now, we need to find V₂. We do this using Charles' law:

$V₁/T₁ = V₂/T₂$

$ V₂ = V₁/T₁ × T₂ = 3 L ÷ 273 K × 110 K$

V₂ = 1.21 L

Now we are in condition to calculate the work done during the isobaric compression:

$W= -P. (V₂ - V₁)$

$W= -22.39 atm. (1.21 - 3) L$

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:

$W = n.R.T. ln (V₂/V₁) $

$W = 3 mol ×0.082 (L.atm/mol.K) × 110 K × ln (3L / 1.21 L) $

W= 24.57 L.atm

We convert this value to joules: W= 2489.58 J