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Thermodynamics First Law Question

  • Thread starter BoanviaFx
  • Start date
1. Homework Statement
Can someone help me confirm if I answered correctly please?

a) A sample of gas is enclosed in a cylinder by a piston. The cylinder is given 225J of energy which expands and pushes the piston 16cm outwards against an atmospheric pressure of 1.01x105Pa.
i) give the equation for the first law of thermodynamics and clearly state the meaning of each symbol used.
ii) Show that for an enclosed fixed mass of gas, the change in work done, ΔW is equal to PΔV where P is the pressure of the gas and ΔV the change in volume.
iii) Assuming that the piston is friction less and has an area of 90cm2, calculate the external work done by the gas and the increase in internal energy of the gas.

b) An ideal gas of density 0.3kgm-3 at 300K and pressure 1.5x105Pa occupies a volume of 1.25x10-3m3. The gas first expands isobarically to a volume of 1.55x10-3m3 and then it is compressed isothermally to its original volume.
i) What is isobaric expansion to its original volume.
ii) Calculate the energy added during the isobaric expansion.
iii) What is the temperature and pressure after the isothermal compression?
iv) Calculate the difference between the initial and final internal energy if the specific heat capacity of the gas at constant volume is 7x102JkgK-1.

2. Homework Equations


3. The Attempt at a Solution

ai) ΔU= ΔQ+ΔW
U is the internal energy of the system, Q is the heat gained or lost by the system and W is the work done by the system or on the system.
ii) W=PΔV
P = F/A = N/m²
V = m3
Si Units cancel out leaving "Nm"
W=F*s
iii) P=F/A
Force acting outwards:
F=PA
F=1.01x105Pa*0.009m²
F=909N

W=Fs
W=909N*0.16m
W=145.44J

bi) Isobaric expansion, is an expansion in an enclosed gas system where the volume increases and the pressure is kept constant.

Isothermal compression involves the internal energy to be constant of an enclosed gas system, therefore we can derive that Q=-W. Heat applied to a system will be converted to work done. This process occurs under constant temperature therefore internal energy will remain constant.

ii) Since the first law of thermodynamics states that:

ΔU= ΔQ+ΔW

Work done: PΔV
W=1.5x105Pa*3x10-4m3
W=45J
Loss in energy due to work done on the surroundings

I'm not sure how to find the change in heat energy, need advice on this one.

iii) Isothermal process takes place with a constant temperature.

So pressure:
p1v1=p2v2
1.5x105Pa*1.55x10-3m3=p2*1.25x10-3m3
p2=186000Pa

During Isobaric expansion temperature wil increase to a point. After Isothermal process takes place the temperature will be the same.

Temperature:
V1/T1=V2/T2
1.25x10-3m3/300K=1.55x10-3m3/T2
T2=372K

v) Again i'm not sure about this one too.
 

ehild

Homework Helper
15,354
1,759
1. Homework Statement
Can someone help me confirm if I answered correctly please?

a) A sample of gas is enclosed in a cylinder by a piston. The cylinder is given 225J of energy which expands and pushes the piston 16cm outwards against an atmospheric pressure of 1.01x105Pa.
i) give the equation for the first law of thermodynamics and clearly state the meaning of each symbol used.
ii) Show that for an enclosed fixed mass of gas, the change in work done, ΔW is equal to PΔV where P is the pressure of the gas and ΔV the change in volume.
iii) Assuming that the piston is friction less and has an area of 90cm2, calculate the external work done by the gas and the increase in internal energy of the gas.

b) An ideal gas of density 0.3kgm-3 at 300K and pressure 1.5x105Pa occupies a volume of 1.25x10-3m3. The gas first expands isobarically to a volume of 1.55x10-3m3 and then it is compressed isothermally to its original volume.
i) What is isobaric expansion to its original volume.
ii) Calculate the energy added during the isobaric expansion.
iii) What is the temperature and pressure after the isothermal compression?
iv) Calculate the difference between the initial and final internal energy if the specific heat capacity of the gas at constant volume is 7x102JkgK-1.

2. Homework Equations


3. The Attempt at a Solution

ai) ΔU= ΔQ+ΔW
U is the internal energy of the system, Q is the heat gained or lost by the system and W is the work done by the system or on the system.
ii) W=PΔV
P = F/A = N/m²
V = m3
Si Units cancel out leaving "Nm"
W=F*s
iii) P=F/A
Force acting outwards:
F=PA
F=1.01x105Pa*0.009m²
F=909N

W=Fs
W=909N*0.16m
W=145.44J
Correct, but use only 3 significant digits in the result.

bi) Isobaric expansion, is an expansion in an enclosed gas system where the volume increases and the pressure is kept constant.

Isothermal compression involves the internal energy to be constant of an enclosed gas system, therefore we can derive that Q=-W. Heat applied to a system will be converted to work done. This process occurs under constant temperature therefore internal energy will remain constant.

ii) Since the first law of thermodynamics states that:

ΔU= ΔQ+ΔW

Work done: PΔV
W=1.5x105Pa*3x10-4m3
W=45J
Loss in energy due to work done on the surroundings

I'm not sure how to find the change in heat energy, need advice on this one.
Using the Ideal Gas Law you can calculate the change of temperature. To get the heat added, use the specific heat capacity of the gas. How is the internal energy related to temperature, amount of gas, and specific heat capacity at constant volume?

iii) Isothermal process takes place with a constant temperature.

So pressure:
p1v1=p2v2
1.5x105Pa*1.55x10-3m3=p2*1.25x10-3m3
p2=186000Pa

During Isobaric expansion temperature wil increase to a point. After Isothermal process takes place the temperature will be the same.

Temperature:
V1/T1=V2/T2
1.25x10-3m3/300K=1.55x10-3m3/T2
T2=372K

v) Again i'm not sure about this one too.
Again, you need how is the internal energy related to heat capacity and temperature.
 
Thanks for your reply! Alright I see what you mean.

Second attempt:

Number of moles in gas using ideal gas equation:
R= 8.314 4621 J mol-1 K-1 (Given in exam paper booklet):

n=(PV/RT)
n=(1.5x105Pa*1.25x10-3m3)/(8.314*300K)
n=0.075

Change in temperature using Charles law would be: 372K

Using: Q=Cv*n*ΔT
Q = 7x102JkgK-1*0.075*72
Q = 3780J

I'm guessing the initial and final internal energy will remain the same since an Isothermal process involves no change change in internal energy.

Also I'm curious if I did something wrong since I did not use the density of the gas anywhere in my calculations.
 

ehild

Homework Helper
15,354
1,759
Thanks for your reply! Alright I see what you mean.

Second attempt:

Number of moles in gas using ideal gas equation:
R= 8.314 4621 J mol-1 K-1 (Given in exam paper booklet):

n=(PV/RT)
n=(1.5x105Pa*1.25x10-3m3)/(8.314*300K)
n=0.075

Change in temperature using Charles law would be: 372K

Using: Q=Cv*n*ΔT
Q = 7x102JkgK-1*0.075*72
Q = 3780J
You are given the specific heat capacity that refers to 1 kg mass. And you use it as the molar heat capacity, which is wrong.
From the given density and volume, you find the mass of the gas. Q=cv*m*ΔT.
 
Thank you so much! That helps a lot. :)
 

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