1. PV diagram below shows an ideal gas undergoes an isothermal process. Calculate the work is done by the gas in the process AB.
Solution
Work done by a gas is equal to the area under the PV curve
AB = triangle area + rectangle area
W = [½ (8 x 105–4 x 105)(3-1)] + [4 x 105 (3-1)]
W = [½ (4 x 105)(2)] + [4 x 105 (2)]
W = [4 x 105] + [8 x 105]
W = 12 x 105 Joule
The work is done by the gas in the process AB = 12 x 105 Joule
[irp]
2. Calculate the work is done by an ideal gas in the process ABC.
The work is done by an ideal gas in the process ABC = the area under the PV curve
AB = triangle area + rectangle area
W = [½(10×105–5×105)(30-10)]+[5×105(30-10)]
W = [½ (5 x 105)(20)] + [5 x 105 (20)]
W = [(5 x 105)(10)] + [100 x 105]
W = [50 x 105] + [100 x 105]
W = 150 x 105 Joule
W = 1.5 x 107 Joule
[irp]
3. An ideal gas undergoing isothermal processes. What is an amount of heat is added to the gas so the gas do work of 5000 Joule on the environment.
Known :
Work (W) = 5000 Joule
Wanted: Heat is added to the gas (Q)
Solution :
An isothermal process is a thermodynamic process that occurs at a constant temperature.
ΔU = 3/2 n R ΔT
ΔU = the change in internal energy, n = number of moles, R = universal gas constant, ΔT = The change in temperature.
According to the above equation, if ΔT = 0 then ΔU = 0.
The equation of the first law of thermodynamics :
ΔU = Q – W
0 = Q – W
Q = W
Q = 5000 Joule.
[irp]
4. PV diagram for an ideal gas undergoing isothermal process shown in the figure below. Calculate the heat is added by a gas in process AB.
Known :
Pressure 1 (P1) = 5 atm = 5 x 105 Pa
Pressure 2 (P2) = 10 atm = 10 x 105 Pa
Volume 1 (V1) = 2 m3
Volume 2 (V2) = 6 m3
Wanted : Heat is added in process AB.
Solution :
Isothermal = constant temperature. According to the equation below, if ΔT = 0 then ΔU = 0.
ΔU = 3/2 n R ΔT
ΔU = 3/2 n R (0)
ΔU = 0
Apply to the first law of thermodynamics :
ΔU = Q-W
0 = Q-W
Q=W
The work is done by an ideal gas = the area under the PV curve = triangle area + rectangle area
W = ½ (P2 – P1)(V2 – V1) + P1 (V2 – V1)
W = ½ (10 x 105 – 5 x 105)(6-2) + (5 x 105)(6-2)
W = ½ (5 x 105)(4) + (5 x 105)(4)
W = ½ (20 x 105) + (20 x 105)
W = (10 x 105) + (20 x 105)
W = 30 x 105 Joule
