Solving Isobaric Expansion Homework Problem - 871.395 J

In summary: But the relationship between pressure and volume is inverse. As the volume increases, the pressure decreases. This is why work is being done. The gas is expanding against the external pressure. Hope this helps.In summary, the problem involves an isobaric expansion of 5.00 g of nitrogen gas at 22.0 deg C and an initial pressure of 2.00 atm. Using the ideal gas law and the first law of thermodynamics, the volume of the gas is found to triple. To calculate the heat energy transferred to the gas, the specific heat at constant pressure (Cp) is used, which is equal to the specific heat at constant volume (Cv) plus the gas constant (R).
  • #1
amdma2003
7
0

Homework Statement


5.00 g of nitrogen gas at 22.0 deg C and an initial pressure of 2.00 atm undergo an isobaric expansion until the volume has tripled.

A. How much heat energy is transferred to the gas to cause this expansion?


Homework Equations



Equation of Isobaric thermodynamic: W = -p(delta V)
Ideal Gas Law: PV=nRT

The Attempt at a Solution


Get the number of moles of the N2 gas: N2 = 28 g/mol ; moles = 5/28
Convert 22 deg C to Kelvin: 273.15 + 22 = 295.15 K
Convert 2.00 atm to Pascals: 2 atm = 202 650 pascals
R is 8.31
Find V initial:
PV = nRT
(202650)V = 0.1786(8.31)295.15
Vi = 0.0022
Find V final: Vf = 3(Vi) = 0.0065

Now Find the Work done:
W = -p(DELTA V)
Since pressure stays constant at 202,650 Pascals then:
W = -202650(0.0065-0.0022) = 871.395 J

However this is wrong, and I do not understand why. Can someone walk me through this problem?
 
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  • #2
A. How much heat energy is transferred to the gas to cause this expansion?

Homework Equations



Equation of Isobaric thermodynamic: W = -p(delta V)
Ideal Gas Law: PV=nRT

The Attempt at a Solution


Get the number of moles of the N2 gas: N2 = 28 g/mol ; moles = 5/28
Convert 22 deg C to Kelvin: 273.15 + 22 = 295.15 K
Convert 2.00 atm to Pascals: 2 atm = 202 650 pascals
R is 8.31
Find V initial:
PV = nRT
(202650)V = 0.1786(8.31)295.15
Vi = 0.0022
Find V final: Vf = 3(Vi) = 0.0065

Now Find the Work done:
W = -p(DELTA V)
Since pressure stays constant at 202,650 Pascals then:
W = -202650(0.0065-0.0022) = 871.395 J

However this is wrong, and I do not understand why. Can someone walk me through this problem?

Start with the first law. Since P is constant:

[tex]\Delta Q = \Delta U + P\Delta V[/tex]

Since [itex]\Delta U = nC_v\Delta T[/itex]

[tex]\Delta Q = nC_v\Delta T + P\Delta V[/tex]

Find the change in T using PV = nRT to find the heat flow [itex]\Delta Q[/itex]

AM
 
Last edited:
  • #3
I'm not familiar with the procedure for this type of question, but it says that 2.00 atm is the initial pressure. I would guess that it would change.

However, as I said, I don't know. Sorry if this is wrong.
 
  • #4
Thanks for directing me to the First Law of Thermodynamics. I have never thought of using the relationship.

However, for the record, I did realize that it should be C_p not C_v as it is the specific heat of the gas at constant pressure.
 
  • #5
amdma2003 said:
Thanks for directing me to the First Law of Thermodynamics. I have never thought of using the relationship.

However, for the record, I did realize that it should be C_p not C_v as it is the specific heat of the gas at constant pressure.
You can certainly use Cp to find the heat flow (Cp takes into account the work done in the expansion):

[tex]\Delta Q = nC_p\Delta T[/tex]

Change in interal energy is ALWAYS [itex]\Delta U = nC_v\Delta T[/itex].

If P is constant:

[tex]\Delta Q = nC_p\Delta T = nC_v\Delta T + P\Delta V[/tex]

But [itex]\Delta V = nR\Delta T/P[/itex] so

[tex]\Delta Q = (nC_v + nR)\Delta T[/tex] so

[tex]C_p = C_v + R[/tex]

AM
 
  • #6
Vidatu said:
I'm not familiar with the procedure for this type of question, but it says that 2.00 atm is the initial pressure. I would guess that it would change.
The problem states "isobaric expansion", so pressure does not change. The volume does change as also stated.
 

1. How do I solve an isobaric expansion homework problem?

To solve an isobaric expansion homework problem, you will need to follow a few steps:
1. Determine the initial and final volume and temperature of the gas in the problem.
2. Use the ideal gas law (PV = nRT) to calculate the number of moles of gas.
3. Apply the equation for isobaric expansion (Q = nCpΔT) to calculate the change in heat.
4. Use the given value for ΔH to solve for the change in internal energy (ΔU).
5. Use the first law of thermodynamics (ΔU = Q + W) to calculate the work done on or by the gas.

2. What is the ideal gas law and how is it used in solving isobaric expansion problems?

The ideal gas law (PV = nRT) is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is used in solving isobaric expansion problems by providing a way to calculate the number of moles of gas present. This is necessary in order to apply the equation for isobaric expansion (Q = nCpΔT) and solve for the change in heat.

3. What is isobaric expansion and how does it affect the internal energy of a gas?

Isobaric expansion is a type of thermodynamic process in which the pressure of a gas remains constant while the volume and temperature change. In this process, work is done on or by the gas, which results in a change in internal energy. If work is done on the gas, the internal energy increases. If work is done by the gas, the internal energy decreases.

4. How do I calculate the change in heat for an isobaric expansion problem?

To calculate the change in heat (Q) for an isobaric expansion problem, you will need to use the equation Q = nCpΔT, where n is the number of moles of gas, Cp is the specific heat capacity of the gas at constant pressure, and ΔT is the change in temperature. The value for Cp can usually be found in a table for common gases.

5. Can I use the first law of thermodynamics to calculate the work done in an isobaric expansion problem?

Yes, you can use the first law of thermodynamics (ΔU = Q + W) to calculate the work done on or by the gas in an isobaric expansion problem. This equation states that the change in internal energy (ΔU) is equal to the sum of the heat added to the system (Q) and the work done on or by the system (W). By rearranging the equation, you can solve for the work done.

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