Week 3 - Part 2

Internal Energy and Ideal Gas

• Recall that Internal Energy is the sum of molecular kinetic and potential energy.

• An ideal gas has no forces acting between molecules - There is no potential energy since:

\[ \vec{F} = -\frac{dU}{dx} \]

• If you were to place an ideal gas in an fixed insulated container and allow it to expand.

  • No heat would be exchanges (insulated)
  • No work would be done (fixed)
  • Volume increases.
  • Pressure decreases.
  • Temperature would remain constant.

• In such experiments preformed with low density gases, Temperature does remain constant.

Non-ideal gases

• In real gases, molecules do apply forces to each other, so there is some potential energy.

• Potential Energy increases with distance.

  • For example, gravity and springs, the greater the height or compression the more PE.

• When real gases expand in an insulated, fixed container, Temperature decreases.

  • More volume means greater distances between molecules.
  • In turn there is more PE.
  • Since internal energy is not changing, KE drops.
  • Temperature measures average KE and drops as well

Heat Capacities of Ideal Gases

• Heat capacity for gases is dependent on the conditions in which the heat is added.

• In changing the temperature of an ideal gas:

  • If volume is held constant, the gas does not do work.

  • If pressure is held constant, the gas does do work.

• So heat is split between internal energy, measured by Temperature in an ideal gas, and work:

\[ Q = \Delta U + W \]

We would expect it would take more heat to change temperature if work is done.

Relating Cp and Cv

• By definition of \(C_V\):

\[ dQ = nC_vdT \\ dU = dQ -dW \\ dW = 0 \\ dU = nC_VdT \]

• By definition for \(C_P\):

\[ dQ = nC_PdT \\ dW = pdV \\ pV = nRT \rightarrow pdV = nRdT\\ dU = dQ -dW \\ dU = nC_PdT - nRdT \]

CP and CV part II

We’ve derived \[ dU = nC_VdT \\ dU = nC_PdT - nRdT \\ nC_PdT = nC_VdT + nRdT \\ \boxed{C_P = C_V + R} \]

• \(C_P\) is always higher than \(C_V\) by adding R.

  • It takes more heat to change temperature when work is done, as expected.

Question

Recall that \(C_V\) for mono- , di- , and polyatomic gases are all factors of R for ideal gases.

What are the corresponding \(C_P\)’s?

Answer

monoatomic:

\[ C_v = \frac{3}{2}R \\ C_P = \frac{5}{2}R \]

diatomic:

\[ C_v = \frac{5}{2}R \\ C_P = \frac{7}{2}R \]

Con’t

polyatomic:

\[ C_v = \frac{7}{2}R \\ C_P = \frac{9}{2}R \]

Heat Capacity Ratios

• In some situations, the ratio of the \(C_V\) and \(C_P\) heat capacities are needed.

• This ratio is called \(\gamma\)

\[ \gamma = \frac{C_P}{C_V} \]

• Question: What are the gammas for the 3 types of ideal gas?

Answer

monoatomic:

\[ \gamma = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} = 1.67 \]

diatomic:

\[ \gamma = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} = 1.4 \]

polyatomic:

\[ \gamma = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} = 1.29 \]

The Adiabatic Processes for Ideal Gas

• The Adiabatic process is an excellent approximation for real process that occur very rapidly, such that the heat exchange is not significant in that time frame.

• Recall that for adiabatic processes:

\[ dU = -W \]

• Temperature changes that occur adiabatically are often referred to as “Adiabatic Heating/Cooling.”

Putting it together

\[ dU = -dW \\ nC_VdT = -pdV \\ nC_VdT = -\frac{-nRT}{V}dV \\ \frac{dT}{T} + \frac{R}{C_V}\frac{dV}{V} =0 \\ \frac{R}{C_V} = \frac{C_P -C_V}{C_V} = \frac{C_P}{C_V} -1 = \gamma -1 \\ \frac{dT}{T} + (\gamma -1) \frac{dV}{V} = 0 \]

Integrating

\[ \int\frac{dT}{T} + \int(\gamma -1) \frac{dV}{V} = \int0 \\ ln(T) + (\gamma -1)ln(V) = constant \\ ln(T) + ln(V^{\gamma -1}) = constant \\ ln(TV^{\gamma -1}) = constant \\ TV^{\gamma -1} = constant \\ \boxed{T_1V_1^{\gamma -1} = T_2V_2^{\gamma -1}} \\ \boxed{p_1V_1^{\gamma} = p_2V_2^{\gamma}} \]

Question 1

1.00 liter of ideal, monoatomic gas at a temperature of 300. K expands adaibatically to a final volume of 100. l, What is the final temperature?

Answer

\[ T_2 = T_1(\frac{V_1}{V_2})^{\gamma-1} \\ T_2 = 300K(\frac{1.00 l}{100l})^{1.67-1} \\ T_2 = 13.9 K \]

Question 2

If the initial pressure in the above question in 1.0 ATM, what is the final pressure?

\[ p_2 = p_1(\frac{V_1}{V_2})^{\gamma} \\ T_2 = 1ATM(\frac{1.00 l}{100l})^{1.67} \\ T_2 = 4.6\times 10^{-4} ATM \]

Adaibatic Work

\[ W = nC_V(T_1 - T_2) \]

Apply \(pV = nRT\)

\[ W = \frac{C_V}{R} (p_1V_1 - p_2V_2) = \frac{1}{\gamma-1}(p_1V_1 - p_2V_2) \]

Question 3

Same gas as the previous two problems. How much work does the gas do?

Answer

\[ W = \frac{1}{\gamma-1}(p_1V_1 - p_2V_2) \\ W = \frac{1}{0.67}(1.01\times10^5 Pa * 0.001 m^3 - 48.88 Pa *0.100 m^3)\\ W =144 J \]