Question

A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environment during each stage of the cycle described below. (Enter your answers in J.)
(a) First, the gas expands from a volume of 1.0 L to 6.0 L at a constant pressure of 6.5 atm.
(b) Second, the gas is cooled at constant volume until the pressure falls to 1.0 atm.
(c) Third, the gas is compressed at a constant pressure of 1.0 atm from a volume of 6.0 L to 1.0 L. (Note: Be careful of signs.)
(d) Finally, the gas is heated until its pressure increases from 1.0 atm to 6.5 atm at a constant volume. (e) What is the net work done by the gas on its environment during the complete cycle described above?

Answer 1

Answer:

(a) W = 3293 J = 3.293 KJ

(b) W = 0 KJ

(c) W = -506.6 J = -0.507 KJ

(d) W = 0 KJ

Net Work = 2.786 KJ = 2786 J

Explanation:

(a)

The work done is given as:

$W = P\Delta V\\W = P(V_2-V_1)$

where,

P = Constant Pressure = (6.5 atm)(101325 Pa/1 atm) = 6.59 x 10⁶ Pa

V₁ = initial volume = (1 L)(0.001 m³/1 L) = 0.001 m³

V₂ = final volume =  (6 L)(0.001 m³/1 L) = 0.006 m³

Therefore,

$W = (6.59\ x\ 10^6\ Pa)(0.006\ m^3-0.001\ m^3)$

W = 3293 J = 3.293 KJ

(b)

Since the volume is constant in this stage. Therefore,

ΔV = 0

As a result:

W = 0 KJ

(c)

The work done is given as:

$W = P\Delta V\\W = P(V_2-V_1)$

where,

P = Constant Pressure = (1 atm)(101325 Pa/1 atm) = 1.01 x 10⁵ Pa

V₁ = initial volume = (6 L)(0.001 m³/1 L) = 0.006 m³

V₂ = final volume =  (1 L)(0.001 m³/1 L) = 0.001 m³

Therefore,

$W = (1.01\ x\ 10^5\ Pa)(0.001\ m^3-0.006\ m^3)$

W = -506.6 J = -0.507 KJ

negative sign show that the work is done on the gas by the environment.

(d)

Since the volume is constant in this stage. Therefore,

ΔV = 0

As a result:

W = 0 KJ

Net work will be the sum of all the works:

Net Work = 3.293 KJ + 0 KJ - 0.507 KJ + 0 KJ

Net Work = 2.786 KJ = 2786 J