# Molecular Specific Heat of an Ideal Gas: Computations

• Const@ntine
In summary, a cylinder with a heavy ram/piston contains air at T = 300 K, Pi = 2.00 * 105 Pa, Vi = 0.350 m3, Mr = 28.9 g/mol, and Cv = 5R/2. The molar heat capacity is needed to find the mass of air inside the cylinder and to calculate the required energy for the temperature to increase from 300 K to 700 K. With a constant volume, M = 28.9 g/mol and Cv = 2,5 * 8,314 J/molK, the molar heat capacity is found to be 719.7 J/kgK. To find the required energy for the temperature
Const@ntine

## Homework Statement

A cylinder with a heavy ram/piston contains air at T = 300 K. Pi = 2.00 * 105 Pa, Vi = 0.350 m3, Mr = 28.9 g/mol & Cv = 5R/2

(a) What's the Molecular Specific Heat of an Ideal Gas, with a constant volume, computed at J/KgC ? (Cv)
(b) What's the mass of the air inside the cylinder?
(c) If the piston stands still, how much energy is required for Tf = 700 K?
(d) If the piston is free to move, how much energy is required for Tf = 700 K?

ΔE = nCvΔT
PV = nRT
Cv = R + Cp

## The Attempt at a Solution

(a) No clue what to do here.

(b) PiVi = nRTi <=> ... <=> n = 28.065 mol = m/Mr = m/28.9(g/mol) <=> ... <=> m = 0.811 kg

(c) ΔE = nCvΔT = 28.065 mol * (5/2 * 8,314 J/molK) * (700 - 300) K <=> ΔE = 233 kJ

(d) No clue what to do here either.

Any help is appreciated!

In part (a) if you know the molar heat capacity (J/mole-K) and you know the molecular weight (gm/mole), what is the heat capacity in J/kg-K?

Chestermiller said:
In part (a) if you know the molar heat capacity (J/mole-K) and you know the molecular weight (gm/mole), what is the heat capacity in J/kg-K?
Kind of a correction here, I made a mistake. (a) is about the Molar Heat Capacity in J/kgC, not the Molecular Specific Heat. So in otherwords c, not Cv or Cp. I'm putting this in bold since I can't edit the OP (kinda messed up here, sorry). The book doesn't explain the differences that well.

Anyway, I couldn't find anything to connect Cv & Cp in my book, but I looked around and found that C = M * c.

So, now we know that the Molecular Specific Heat of an Ideal Gas and the Molar Heat Capacity are connected through this formula: C = M * c

In this case we have a constant volume and M = 28.9 g/mol. We are given that Cv = 5/2*R = 2,5 * 8,314 J/molK

20,785 J/molK = 0,0289 kg/mol * c <=> c = 719.7 J/kgK. So now I need to somehow turn the K into a C.

Darthkostis said:
Kind of a correction here, I made a mistake. (a) is about the Molar Heat Capacity in J/kgC, not the Molecular Specific Heat. So in otherwords c, not Cv or Cp. I'm putting this in bold since I can't edit the OP (kinda messed up here, sorry). The book doesn't explain the differences that well.

Anyway, I couldn't find anything to connect Cv & Cp in my book, but I looked around and found that C = M * c.

So, now we know that the Molecular Specific Heat of an Ideal Gas and the Molar Heat Capacity are connected through this formula: C = M * c

In this case we have a constant volume and M = 28.9 g/mol. We are given that Cv = 5/2*R = 2,5 * 8,314 J/molK

20,785 J/molK = 0,0289 kg/mol * c <=> c = 719.7 J/kgK. So now I need to somehow turn the K into a C.
The molar heat capacity refers to J/molC, not J/kgC. I, who have had lots of experience with thermodynamics, am having trouble interpreting the terminology you are using.

The equation C = Mc is correct, but I'm surprised you had to refer to your book for this. Have you learned to work with cancellation of units?

You are aware that a change of 1 K is the same as a change of 1 C?

Chestermiller said:
The molar heat capacity refers to J/molC, not J/kgC.

Yeah, I made a mistake in the OP.

Chestermiller said:
I, who have had lots of experience with thermodynamics, am having trouble interpreting the terminology you are using.

Well, english isn't my native, which means that my books are in another tongue. So when I want to ask something I have to search online for the correct words in order to translate the exercises, and because I do not know the proper terminology I get a bit lost sometimes.

Chestermiller said:
The equation C = Mc is correct, but I'm surprised you had to refer to your book for this.

The theory wasn't that clear on the distinction between the two, so I'm still a bit iffy on it.

Chestermiller said:
Have you learned to work with cancellation of units?

You mean stuff like 1 kg = 103 g & 1 atm = 101325 Pa, right?

Chestermiller said:
You are aware that a change of 1 K is the same as a change of 1 C?

Yes, that's in my book (in the "end notes" section). Does that mean that I can simply "replace" K with C in my c = 719.7 J/kgK result?

Darthkostis said:
Yes, that's in my book (in the "end notes" section). Does that mean that I can simply "replace" K with C in my c = 719.7 J/kgK result?
Yes.

Chestermiller said:
Yes.
Okay, gotcha.

As for (d), this is what I came up with:

Q = mcΔT = nMrcΔT = ... = 233 kJ

P = constant, so: PfVf = nRTf <=> Vf = 8.065 mol * 8,314 J/molK * 700 K / 2*105 Pa = 0.820 m3

W = -P∫VfVidV = ... = - 94000 J

Now, the Work is negative, which means that since the gas' volume expands, it's energy from the gas that is applied to the piston. Either way, (d) asks for the total energy that is required for T to go from 300 K to 700 K, so I assume it means all the energies, regardless of the "direction". So:

ΔΕ = Q + |W| = 327 kJ

This is the book's answer, but I'm not sure whethermy method is correct. The exercise just says that "the initial data is valid, and the piston can move freely".

Darthkostis said:
Okay, gotcha.

As for (d), this is what I came up with:

Q = mcΔT = nMrcΔT = ... = 233 kJ

P = constant, so: PfVf = nRTf <=> Vf = 8.065 mol * 8,314 J/molK * 700 K / 2*105 Pa = 0.820 m3

W = -P∫VfVidV = ... = - 94000 J

Now, the Work is negative, which means that since the gas' volume expands, it's energy from the gas that is applied to the piston. Either way, (d) asks for the total energy that is required for T to go from 300 K to 700 K, so I assume it means all the energies, regardless of the "direction". So:

ΔΕ = Q + |W| = 327 kJ

This is the book's answer, but I'm not sure whethermy method is correct. The exercise just says that "the initial data is valid, and the piston can move freely".
Which version of the first law are you using, $$\Delta E = Q - W$$where W is the work done by the gas on the piston or $$\Delta E=Q+W$$where W is the work done by the piston on the gas?

Also, in the last part (d), you are not trying to determine ##\Delta E##. You are trying to determine Q.

Const@ntine
Chestermiller said:
Which version of the first law are you using, $$\Delta E = Q - W$$where W is the work done by the gas on the piston or $$\Delta E=Q+W$$where W is the work done by the piston on the gas?

I'm going with the second. I know about the two versions, but it's the one I've been taught since HS. Anyway, with my theory we have: W > 0 means that it's work done on the gas, W < 0 means it's work done by the gas.

Chestermiller said:
Also, in the last part (d), you are not trying to determine ##\Delta E##. You are trying to determine Q.

Oh yeah, no wonder I had to do "mental gymnastics" to get the correct result. My bad. It should be:

ΔE = nCvΔT = ... = 233 kJ
W = -94000 J (as shown in my above post-the n should be 28.065 up there, my keyboard's a bit busted)

So: ΔE = Q + W <=> 233 kJ = Q - 94 kJ <=> Q = 327 kJ

PS: Thanks a ton for the help!

Chestermiller

## What is the molecular specific heat of an ideal gas?

The molecular specific heat of an ideal gas is a measure of the amount of heat energy needed to raise the temperature of one mole of the gas by one degree Celsius. It is represented by the symbol Cv and is a characteristic property of each gas molecule.

## How is the molecular specific heat of an ideal gas calculated?

The molecular specific heat of an ideal gas can be calculated using the equation Cv = dQ / dT, where dQ is the change in heat energy and dT is the change in temperature. This calculation can also be done using the molar specific heat capacity, which is the heat energy needed to raise the temperature of one mole of the gas by one degree Celsius.

## What factors affect the molecular specific heat of an ideal gas?

The molecular specific heat of an ideal gas can be affected by factors such as the number of atoms in the gas molecule, the type of atoms in the molecule, and the temperature and pressure of the gas. It may also vary depending on the conditions under which the gas is measured, such as constant volume or constant pressure.

## Why is the molecular specific heat of an ideal gas important?

The molecular specific heat of an ideal gas is an important property in understanding and predicting the behavior of gases. It is used in thermodynamic calculations and can help determine the efficiency of heat engines, as well as in designing and optimizing industrial processes involving gases.

## How does the molecular specific heat of an ideal gas differ from that of a real gas?

An ideal gas is a theoretical concept that follows the gas laws perfectly, while a real gas deviates from these laws due to intermolecular forces and molecular size. Therefore, the molecular specific heat of an ideal gas and a real gas may differ, with real gases typically having higher values due to the added factors of intermolecular interactions and molecular structure.

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