1. Ideal gases are in a container with a volume of 4 liters and its pressure is 3 atm (1 atm = 105 N.m-2). The ideal gases heated at a constant pressure from 27oC to 87oC. The heat capacity of the gas is 9 J.K-1. What is the final volume of gases and the change of internal energy of gases?
Solution
Isobaric process (constant pressure)
Known :
The initial volume of gas (V1) = 4 liters
The initial temperature of gas (T1) = 27oC + 273 = 300 K
The final temperature of gas (T2) = 87oC + 273 = 360 K
The pressure of gas (P) = 3 atm = 3 x 105 N.m-2
The heat capacity of gas (C) = 9 J.K-1
Wanted : The final volume of gas (V2) and the change of internal energy of gas (ΔU)
Solution :
Determine the final volume using the equation of Charles’s law (isobaric process or constant pressure) :
The change in volume :
1 liter = 0.001 m3
Initial volume (V1) = 4 (0.001 m3) = 0.004 m3
Final volume (V2) = 4.8 (0.001 m3) = 0.0048 m3
The change in volume (ΔV) = V2 – V1 = 0.0048 m3 – 0.004 m3 = 0.008 m3.
The change in temperature :
The change in temperature (ΔT) = T2 – T1 = 360 K – 300 K = 60 K
Determine the change of the internal energy (ΔU) of the ideal gas using the equation of the first law of thermodynamics.
ΔU = Q – W
ΔU = the change of the internal energy, Q = heat, W = work
Determine work (W) at constant pressure :
W = P ΔV = (3 x 105)(0.0008) = (3 x 101)(8) = (30)(8) = 240 Joule
Determine heat (Q) using the equation of the heat capacity (C) :
C = Q / ΔT
Q = (C)(ΔT) = (9)(60) = 540 Joule
Determine the change of the internal energy :
ΔU = Q – W = 540 Joule – 240 Joule = 300 Joule.
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2. 6 liters of ideal gases at 2 atm are in a container (1 atm = 105 N.m-2). The gas heated from 27oC to 77oC at a constant pressure. If the heat capacity of gas is 5 J.K-1, what is the final volume and the change of internal energy of the gas.
Solution :
Isobaric process (constant pressure)
Known :
The initial volume of the ideal gases (V1) = 6 liters
The initial temperature of the ideal gases (T1) = 27oC + 273 = 300 K
The final temperature of the ideal gases (T2) = 77oC + 273 = 350 K
The pressure of the ideal gases (P) = 2 atm = 2 x 105 N.m-2
The heat capacity of gases (C) = 5 J.K-1
Wanted: The final volume of the gas (V2) and the change of internal energy of the gas (ΔU)
Solution :
Determine the final volume of gas using the equation of Charles’s law (isobaric process or constant pressure) :
The change in volume :
1 liter = 0.001 m3
The initial volume (V1) = 6 (0.001 m3) = 0.006 m3
The final volume (V2) = 7 (0.001 m3) = 0.007 m3
The change in volume (ΔV) = V2 – V1 = 0.007 m3 – 0.006 m3 = 0.001 m3
The change in temperature :
The change in temperature (ΔT) = T2 – T1 = 350 K – 300 K = 50 K
Determine the change of the internal energy (ΔU) of the ideal gases using the equation of the first law of thermodynamics.
ΔU = Q – W
ΔU = the change in the internal energy, Q = heat, W = work
Determine work (W) at constant pressure :
W = P ΔV = (2 x 105)(0.001) = (2 x 102)(1) = (200)(1) = 200 Joule
Determine heat (Q) using the equation of the heat capacity (C) :
C = Q / ΔT
Q = (C)(ΔT) = (5)(50) = 250 Joule
Determine the change in the internal energy :
ΔU = Q – W = 250 Joule – 200 Joule = 50 Joule.
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