Thermodynamics - temperature, pressure and heat

[BobaJ]

Homework Statement

The initial state of 0.1 mol of an ideal monatomic gas is P₀=32 Pa and v₀=8m³. The final state is P₁=1 Pa and V₁=64m³. Suppose that the gas undergoes a process along a straight line joining these two states with an equation P=aV+b, where a =31/56 and b=255/7. Plot this straight line to scale on a PV diagram.
Calculate:
a) Temperature T as a function of V along the straight line.
b) The value of V which T is a maximum.
c) The values of T₀, T_max and T₁.
d) The heat Q transferred from te Volume V₀ to any other volume V along the straight line.
e) The values of P and V at which Q is a maximum.
f) The heat transferred along the line from V₀ to V when Q is a maximum.
g) The heat transferred from V at maximum Q to V₁.

Homework Equations

It's a monatomic gas, so γ=5/3.

The Attempt at a Solution

I have already solved a), b) and c).

a) $$T=\frac{1}{nR}*(aV^2+b)$$
b) Take the first derivate of the last result and equal it to 0 $$V=32.9 m^3$$
c) Just insert the desires values of V in the equation for T:
$$T_{0}=307.9 K$$
$$T_{max} = 720.8 K$$
$$T_{1} = 76.97 K$$

So, now I'm stuck on point d). For a moment I thought I could just take $$Q = \int^V_{V_{0}} P dV $$ and insert the given equation for P. But I'm not sure.

Thanks for your help.

 

[Chestermiller]

Homework Statement

The initial state of 0.1 mol of an ideal monatomic gas is P₀=32 Pa and v₀=8m³. The final state is P₁=1 Pa and V₁=64m³. Suppose that the gas undergoes a process along a straight line joining these two states with an equation P=aV+b, where a =31/56 and b=255/7. Plot this straight line to scale on a PV diagram.
Calculate:
a) Temperature T as a function of V along the straight line.
b) The value of V which T is a maximum.
c) The values of T₀, T_max and T₁.
d) The heat Q transferred from te Volume V₀ to any other volume V along the straight line.
e) The values of P and V at which Q is a maximum.
f) The heat transferred along the line from V₀ to V when Q is a maximum.
g) The heat transferred from V at maximum Q to V₁.

Homework Equations

It's a monatomic gas, so γ=5/3.

The Attempt at a Solution

I have already solved a), b) and c).

a) $$T=\frac{1}{nR}*(aV^2+b)$$
b) Take the first derivate of the last result and equal it to 0 $$V=32.9 m^3$$
c) Just insert the desires values of V in the equation for T:
$$T_{0}=307.9 K$$
$$T_{max} = 720.8 K$$
$$T_{1} = 76.97 K$$

So, now I'm stuck on point d). For a moment I thought I could just take $$Q = \int^V_{V_{0}} P dV $$ and insert the given equation for P. But I'm not sure.

Thanks for your help.

What is your rationale for that last equation?

 

[BobaJ]

What is your rationale for that last equation?

Well, honestly I thought I could use it, because it relates the variables that I need, but I can't really justify it.

 

[Chestermiller]

Well, honestly I thought I could use it, because it relates the variables that I need, but I can't really justify it.

Just a wild guess: You're currently learning about the 1st law of thermodynamics, correct. If so, please write down your equation for the 1st law of thermodynamics.

 

[Chestermiller]

Homework Statement

The initial state of 0.1 mol of an ideal monatomic gas is P₀=32 Pa and v₀=8m³. The final state is P₁=1 Pa and V₁=64m³. Suppose that the gas undergoes a process along a straight line joining these two states with an equation P=aV+b, where a =31/56 and b=255/7. Plot this straight line to scale on a PV diagram.
Calculate:
a) Temperature T as a function of V along the straight line.
b) The value of V which T is a maximum.
c) The values of T₀, T_max and T₁.
d) The heat Q transferred from te Volume V₀ to any other volume V along the straight line.
e) The values of P and V at which Q is a maximum.
f) The heat transferred along the line from V₀ to V when Q is a maximum.
g) The heat transferred from V at maximum Q to V₁.

Homework Equations

It's a monatomic gas, so γ=5/3.

The Attempt at a Solution

I have already solved a), b) and c).

a) $$T=\frac{1}{nR}*(aV^2+b)$$

Shouldn't this equation read: $$T=\frac{V(aV+b)}{nR}$$

 

[Chestermiller]

Homework Statement

The initial state of 0.1 mol of an ideal monatomic gas is P₀=32 Pa and v₀=8m³. The final state is P₁=1 Pa and V₁=64m³. Suppose that the gas undergoes a process along a straight line joining these two states with an equation P=aV+b, where a =31/56 and b=255/7. Plot this straight line to scale on a PV diagram.
Calculate:
a) Temperature T as a function of V along the straight line.
b) The value of V which T is a maximum.
c) The values of T₀, T_max and T₁.
d) The heat Q transferred from te Volume V₀ to any other volume V along the straight line.
e) The values of P and V at which Q is a maximum.
f) The heat transferred along the line from V₀ to V when Q is a maximum.
g) The heat transferred from V at maximum Q to V₁.

Homework Equations

It's a monatomic gas, so γ=5/3.

The Attempt at a Solution

I have already solved a), b) and c).

a) $$T=\frac{1}{nR}*(aV^2+b)$$
b) Take the first derivate of the last result and equal it to 0 $$V=32.9 m^3$$
c) Just insert the desires values of V in the equation for T:
$$T_{0}=307.9 K$$
$$T_{max} = 720.8 K$$
$$T_{1} = 76.97 K$$

So, now I'm stuck on point d). For a moment I thought I could just take $$Q = \int^V_{V_{0}} P dV $$ and insert the given equation for P. But I'm not sure.

Thanks for your help.

The parameter a should be -31/56, not +31/56. On physics problems, the hard part is the physics, and the simple part is supposed to be the math. The math should be a gimme. You can't get thermodynamics problems correct if you mess up on the math.

 

[BobaJ]

Shouldn't this equation read: $$T=\frac{V(aV+b)}{nR}$$

The parameter a should be -31/56, not +31/56. On physics problems, the hard part is the physics, and the simple part is supposed to be the math. The math should be a gimme. You can't get thermodynamics problems correct if you mess up on the math.

Yes, you are absolutely right, both where typing mistakes I made. I'm sorry.

The first law of thermodynamics would be: $$dU=dQ+dW$$

 

[Chestermiller]

Yes, you are absolutely right, both where typing mistakes I made. I'm sorry.

The first law of thermodynamics would be: $$dU=dQ+dW$$

So, dW is the work done by the surroundings on the system: dW=-PdV, right? For an ideal monatomic gas, what is the equation for dU in terms of dT? Do you see what you omitted from your analysis now?

 

[BobaJ]

ok, so we would have $$dU=dQ-P dV$$, so $$dQ = dU +P dV$$.
And if I'm not wrong $$dU = C_{V}dT$$. Substituting that would give:
$$dQ=C_{V}dT+PdV$$
So, basically I omitted the part of Cv dT.
Is that right?

 

[Chestermiller]

ok, so we would have $$dU=dQ-P dV$$, so $$dQ = dU +P dV$$.
And if I'm not wrong $$dU = C_{V}dT$$. Substituting that would give:
$$dQ=C_{V}dT+PdV$$
So, basically I omitted the part of Cv dT.
Is that right?

Very nice. But don't forget the n in $dU=nC_vdT$. So now, using your equation for T vs V, what is dT in terms of dV, and what is dU? Then, what is dQ in terms of dV?

 

[BobaJ]

Very nice. But don't forget the n in $dU=nC_vdT$. So now, using your equation for T vs V, what is dT in terms of dV, and what is dU? Then, what is dQ in terms of dV?

Using the result of a): $$dT=\frac{2aV+b}{nR} dV$$ and P=aV+b
As we are working with a monatomic gas: Cv=3/2R.

Putting this into the equation for dQ:
$$dQ=\frac{3}{2}(2aV+b)+(aV+b) dV$$
So,
$$dQ=(4aV+\frac{5}{2}b) dV$$

Am I correct up to this point?

After this, we would have to integrate. But do I have to integrate from V0 to V?

 

[Chestermiller]

Using the result of a): $$dT=\frac{2aV+b}{nR} dV$$ and P=aV+b
As we are working with a monatomic gas: Cv=3/2R.

Putting this into the equation for dQ:
$$dQ=\frac{3}{2}(2aV+b)+(aV+b) dV$$
So,
$$dQ=(4aV+\frac{5}{2}b) dV$$

Am I correct up to this point?

Yes.

After this, we would have to integrate. But do I have to integrate from V0 to V?

Yes. That would give you the answer to part (d).

Now, for part (e), at what value of V is Q maximum?