Solving Isobaric Expansion Homework Problem - 871.395 J

[amdma2003]

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?

 

[Andrew Mason]

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:

$$\Delta Q = \Delta U + P\Delta V$$

Since $\Delta U = nC_v\Delta T$

$$\Delta Q = nC_v\Delta T + P\Delta V$$

Find the change in T using PV = nRT to find the heat flow $\Delta Q$

AM

 

[Vidatu]

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.

 

[amdma2003]

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.

 

[Andrew Mason]

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):

$$\Delta Q = nC_p\Delta T$$

Change in interal energy is ALWAYS $\Delta U = nC_v\Delta T$.

If P is constant:

$$\Delta Q = nC_p\Delta T = nC_v\Delta T + P\Delta V$$

But $\Delta V = nR\Delta T/P$ so

$$\Delta Q = (nC_v + nR)\Delta T$$ so

$$C_p = C_v + R$$

AM

 

[Astronuc]

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.