Change in internal Energy During Expansion of an Ideal Gas

[Purple Baron]

Homework Statement

Derive an expression for the change in internal energy of 1 mol of an ideal gas for expansion from volume V_i to V_f under constant pressure of 1atm where \gamma = \frac{C_p}{C_v}=\frac{5}{3}

Homework Equations

\Delta U=\frac{3}{2}nR\Delta T
PV=nRT

The Attempt at a Solution

My first idea is to find a way to get the change in temperature, I tried to do this by finding the initial temperature saying that: T_i=\frac{PV_i}{nR} and finding the final temperature by substituting this into T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i and substituting the equation obtained for initial temperature. From this I found change in temperature and substituted into the change in internal energy equation did the algebra and obtained \Delta U=\frac{3}{2}PV_i((\frac{V_i}{V_f})^{\gamma - 1}-1); however I'm not sure about the answer as the question also says you may use the fact that for 1 mol of an ideal gas C_p-C_v=R however nowhere did I need to use this, indeed, R canceled out of my final expression also th internal energy in my expression is dependent on the pressure, which I know is not the case, so I am unsure this is a valid approach. Thanks.

 

[Chestermiller]

This problem is way underspecified. They don't tell you whether the expansion takes place isothermally or adiabatically. They don't tell you whether the external pressure is initially higher than 1 atm (matching the initial gas pressure), and is then suddenly dropped to 1 atm, and held at that value until the gas stops expanding on its own, or whether the expansion is stopped before that.

The equation you employed assumes an adiabatic reversible expansion, but, if you check the pressure along the way using the ideal gas law, you will find that the only time the pressure is 1 atm is at the initial condition. So, it can't be an adiabatic reversible expansion.

The only thing you know for sure from the problem statement is that the external pressure is held constant at 1 atm. during the expansion. Since the work done on the surroundings is $W = \int{P_{ext}dV}$, you can say that $W = P_{ext}(V_f-V_i)$, where P_ext is 1 atm.

I also think that they intended for you to assume that the expansion takes place adiabatically, since, if it took place isothermally, the change in internal energy would be zero. I also think that they intended for you to assume that external pressure was initially higher than 1 atm (matching the initial gas pressure), and then was dropped suddenly to 1 atm., and held at that value until the gas stopped expanding on its own. But, we don't know either of these things for sure.

Chet

 

[TSny]

I tried to do this by finding the initial temperature saying that: T_i=\frac{PV_i}{nR} and finding the final temperature by substituting this into T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i and substituting the equation obtained for initial temperature.

The equation T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i is for an adiabatic process. Your exercise is for a constant pressure process.

You need the temperature change. You already have T_i=\frac{PV_i}{nR}. Can you write a very similar expression for $T_f$ in terms of $V_f$?

 

[Quantum Defect]

Homework Statement

Derive an expression for the change in internal energy of 1 mol of an ideal gas for expansion from volume V_i to V_f under constant pressure of 1atm where \gamma = \frac{C_p}{C_v}=\frac{5}{3}

Homework Equations

\Delta U=\frac{3}{2}nR\Delta T
PV=nRT

The Attempt at a Solution

My first idea is to find a way to get the change in temperature, I tried to do this by finding the initial temperature saying that: T_i=\frac{PV_i}{nR} and finding the final temperature by substituting this into T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i and substituting the equation obtained for initial temperature. From this I found change in temperature and substituted into the change in internal energy equation did the algebra and obtained \Delta U=\frac{3}{2}PV_i((\frac{V_i}{V_f})^{\gamma - 1}-1); however I'm not sure about the answer as the question also says you may use the fact that for 1 mol of an ideal gas C_p-C_v=R however nowhere did I need to use this, indeed, R canceled out of my final expression also th internal energy in my expression is dependent on the pressure, which I know is not the case, so I am unsure this is a valid approach. Thanks.

For an ideal gas, the inernal energy is a function of temperature only.

You know Pi, Vi, n, Pf, Vf (Pi = Pf = 1 atm; n = 1 mol). You need an expression for Delta T in terms of Vi, Vf, and the constants (n, R, P) which you can then plug into the expression for Delta U.

[You did indeed use the fact that gamma = 5/3 to determine a value for C_v -- that you used in the expression for Delta U. ]

 

[Chestermiller]

The equation T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i is for an adiabatic process. Your exercise is for a constant pressure process.

You need the temperature change. You already have T_i=\frac{PV_i}{nR}. Can you write a very similar expression for $T_f$ in terms of $V_f$?

Ah. I think TSny has nailed it. They want you to assume that the external pressure is held constant (at 1 atm) while you add heat to make the gas expand (starting and ending at 1 atm pressure). So, it's neither adiabatic nor isothermal. That makes sense.

Chet

 

[Purple Baron]

The equation T_f=\frac{V_{i}^{\gamma - 1}}{V_{f}^{\gamma - 1}}T_i is for an adiabatic process. Your exercise is for a constant pressure process.

You need the temperature change. You already have T_i=\frac{PV_i}{nR}. Can you write a very similar expression for $T_f$ in terms of $V_f$?

Well as it is constant pressure I could use the equation T_f=\frac{T_i V_f}{V_i} substituting initial temperature from the other equation gives T_f=\frac{PV_f}{R} and substitution into the internal energy equation gives \Delta U=\frac{3}{2}P(V_f-V_i) Is this ok?

Thanks for the help everyone.

 

[TSny]

T_f=\frac{PV_f}{R}

The number of moles should appear in this expression somewhere. Note that another way to obtain the expression is just by rearranging the ideal gas law for T.

..substitution into the internal energy equation gives \Delta U=\frac{3}{2}P(V_f-V_i) Is this ok?

Looks good.