A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01m^3. A constant pressure process gives 54 kJ of work out. Find the final volume of the air.
0 points
1)Option 1 0.10 m^3
2)Option 2 0.01 m^3
Answers
Answer:
5 moles of gas in a cylinder undergo an isobaric expansion starting at 293 K. The cylinder is initially 50 cm tall. The radius of the cylinder is 10 cm and Δy is 1 cm. How much work is done by the gas?
Solution:
Concepts:
Isobaric expansion, the ideal gas law
Reasoning:
The process is isobaric, the pressure P is constant. If the volume increase from Vinitial to Vfinal, then the work done by the gas is W = P(Vfinal - Vinitial)
Details of the calculation:
W = PΔV = (nRT/V)ΔV = (5*8.31*293/(0.5 π 0.12) )*(0.01 π 0.12) J = 243 J
Problem:
One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1 atm and volume of 25 L. Determine
(a) the initial volume and
(b) the temperature of the gas.
Solution:
Concepts:
Ideal gas law: PV = nRT, work done on the system: W = -∫PdV
Energy conservation: ΔU = ΔQ + ΔW
Reasoning:
For an isothermal process W = nRT ln(Vi/Vf).
We are given n, Pfinal, and Vfinal, and we are told that the temperature is constant.
Details of the calculation:
(a) For an isothermal process the temperature is constant.
Therefore PV = nRT = constant.
PV = 101000 Pa*25*10-3 m3 = nRT
W = nRT ln(Vi/Vf) for an isothermal process.
W/(nRT) = -3000 J/(101000 Pa*25*10-3 m3) = -1.19
(The work done by the gas is -W.)
Vf/Vi = exp(1.19) = 3.28
Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L
(b) For an ideal gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K)T.
T = 303.85 K.
Problem:
A cylinder contains He gas. In its initial state a the cylinder Volume is Va = 4.23*10-3 m3 and the pressure is Pa = 1.19*10-5 Pa. The volume is isothermically reduced until the system reaches a state b with Vb = 0.581*10-3 m3. This process is followed by an isobaric expansion which allows the system to reach a state c with Vc = Va. The cycle finishes with an isochoric or isometric (constant volume) process that brings the system back to state a.
(a) Draw the cycle in a P-V diagram.
(b) Find the work done by the system going from b to c.
(c) Find the work done by the system in one cycle.
(d) Find the fractional change of the temperature of the system going from b to c.
(e) When the system goes from c to a, does it absorb or release heat? Find the amount of heat absorbed or released.
(f) When the system goes from b to c, does it absorb or release heat? Find the amount of heat absorbed or released.
Solution:
Concepts:
Ideal gas law: PV = nRT, work done on the system: W = -∫PdV
Energy conservation: ΔU = ΔQ - ΔW
increase in internal energy of a system = heat put into the system - work done by the system on its surroundings,
Reasoning:
An isothermal process occurs at constant temperature. P = nRT/V . Since the internal energy of a gas is only a function of its temperature, ΔU = 0 for an isothermal process. For the isothermal expansion of an ideal gas we have
W = ∫V1V2PdV = ∫V1V2nRT dV/V = nRT∫V1V2 dV/V = nRT ln(V2/V1).
W is negative if V2 < V1. Since ΔU = 0, the heat transferred to the gas is ΔQ = W.
An isobaric is a process that occurs at constant pressure. We then have
W = P∫V1V2 dV = P(V2 - V1).
If the pressure of an ideal gas is kept constant, then the temperature must increase as the gas expands. (PV/T = constant.) Heat must be added during the expansion process.
An isometric process takes place at constant volume. Then W = 0 and ΔU = ΔQ. All the heat added to the system goes into increasing its internal energy.
Option(1) 0.1 m³ is the correct answer.
Given,
Pressure(P) = 600kPa
Initial volume (V1) = 0.01 m³
Temperature = 290k
Work done(w) = 54kJ
To Find,
The final volume of the air =?
Solution,
From the formula of work done we know that,
w = PΔV
where w is work done, P is pressure and ΔV is change in volume
54kJ = 600kPa x ΔV
ΔV = 54 / 600
ΔV = 0.09 m³
ΔV = Final volume - Initial volume = 0.09 m³
Final volume = 0.09 m³ + 0.01 m³
Final volume = 0.1 m³
Hence, the final volume of the air is 0.1 m³.
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