In two separate experiments, a sample of helium gas is taken from the same initial state to the same final state, along two separates paths on a P-V diagram. Path a has two parts: first an isochor and second isobar. Path b is an isotherm. Determine the work done by the gas (i) along path a (ii) along path b. The pressure and volume at the initial and final state are Pi=200 KPa, Vi =42 L, Pf =90 KPa and Vf =100L.
To determine the work done by the gas along paths a and b, we first need to understand the equations and principles related to work in thermodynamics.
The general equation for work done by or on a gas is given by:
W = ∫P dV
Here,
W represents the work done by the gas
P represents the pressure
dV represents a small change in volume
In this case, since we have a P-V diagram and information about the initial and final states, we can calculate the work done by integrating the pressure with respect to the volume along each path.
Now let's calculate the work done along each path separately:
(i) Path a: Isochor (constant volume) followed by an isobar (constant pressure)
Step 1: Calculating the work done during the isochor:
Since the volume is constant during the isochor process (V1 = V2), the work done is zero, as there is no change in volume.
Step 2: Calculating the work done during the isobar:
The work done during an isobar is given by:
W = P * (Vf - Vi)
Given: Pi = 200 KPa, Pf = 90 KPa, Vi = 42 L, Vf = 100 L
W = Pf * (Vf - Vi)
Substituting the values:
W = 90 KPa * (100 L - 42 L)
W = 90 KPa * 58 L
Thus, the work done along path a is 5220 KJ (Kilojoules) or 5.22 MJ (Megajoules).
(ii) Path b: Isotherm (constant temperature)
On an isotherm, the work done can be calculated using the formula:
W = nRT * ln(V2/V1)
Where:
n represents the number of moles of gas
R is the ideal gas constant
T is the temperature (in Kelvin)
V1 and V2 are the initial and final volumes, respectively
Since the question does not provide the number of moles, we cannot directly calculate the work done along path b without this information. The work done depends on the number of moles of the gas.
Therefore, we cannot determine the specific value of the work done along path b without further information.
To determine the work done by the gas along the two paths, we can use the formula:
Work = ∫PdV
(i) Along path a:
Path a consists of two parts - an isochor (constant volume) and an isobar (constant pressure).
1. Isochoric process:
In an isochoric process, the volume remains constant, so the work done is zero, as dV = 0.
2. Isobaric process:
Here, the pressure remains constant at Pi = 200 kPa.
Using the equation for work, we have:
Work = ∫PdV = PΔV
Given:
Pi = 200 kPa
Vi = 42 L
Pf = 90 kPa
Vf = 100 L
Using the formula PΔV:
Work = (Pf - Pi) * (Vf - Vi)
Work along path a = (90 kPa - 200 kPa) * (100 L - 42 L)
= (-110 kPa) * (58 L)
= -6380 kPa·L (Negative sign indicates work done by the gas.)
(ii) Along path b:
Path b is an isotherm, meaning the temperature remains constant during the process.
Using the ideal gas law, PV = nRT, where n is the number of moles and R is the gas constant, we can rearrange the equation to:
P = nRT/V
As the temperature is constant, the pressure is inversely proportional to the volume.
Since the path is an isotherm, the work done is equal to the area under the curve on the P-V diagram.
Area under the isotherm = 0.5*(Pf+Pi)*(Vf-Vi)
Using the given values:
Work along path b = 0.5*(90 kPa + 200 kPa)*(100 L - 42 L)
= 0.5*(290 kPa)*(58 L)
= 8410 kPa·L
Therefore, the work done by the gas is:
(i) Along path a: -6380 kPa·L
(ii) Along path b: 8410 kPa·L