Discrepancy regarding a thermodynamics question

[Apashanka]

Homework Statement

Homework Equations

For this problem the correct option is given (c)

The Attempt at a Solution

If p is the final pressure after the partition is removed then p(V₁+V₂)=(n₁+n₂)RT_f
T_f=p(V₁+V₂)/[R(n₁+n₂)]
If n₁ and n₂ moles of the ideal gas is taken in the two compartments.

now regarding denominator n₁R=p₁V₁/T₁ and similarly.
but for the numerator to satisfy option (c)
p=p₁/p=p₂,here is the discrepancy??

 

[stockzahn]

Homework Statement

View attachment 237437

Homework Equations

For this problem the correct option is given (c)

The Attempt at a Solution

If p is the final pressure after the partition is removed then p(V₁+V₂)=(n₁+n₂)RT_f
T_f=p(V₁+V₂)/[R(n₁+n₂)]
If n₁ and n₂ moles of the ideal gas is taken in the two compartments.

now regarding denominator n₁R=p₁V₁/T₁ and similarly.
but for the numerator to satisfy option (c)
p=p₁/p=p₂,here is the discrepancy??

$p\left(V_1+V_2\right)=p_1V_1+p_2V_2 \rightarrow V_1\left(p-p_1\right)=V_2\left(p_2-p\right)$ doesn't mean necessarily $p=p_1=p_2$

 

[Apashanka]

$p\left(V_1+V_2\right)=p_1V_1+p_2V_2 \rightarrow V_1\left(p-p_1\right)=V_2\left(p_2-p\right)$ doesn't mean necessarily $p=p_1=p_2$

Okk if assuming that correct option is not given then how to arrive from

Tf=p(V1+V2)/[R(n1+n2)]

to (p₁V₁+p₂V₂/(n₁R+n₂R)

 

[stockzahn]

Okk if assuming that correct option is not given then how to arrive from

to (p₁V₁+p₂V₂/(n₁R+n₂R)

Energy conservation, assuming that $c_{mv}=const.$:

$n_1c_{mv}T_1+n_2c_{mv}T_2=n_{tot}c_{mv}T_f$

 

[Apashanka]

Energy conservation, assuming that $c_{mv}=const.$:

$n_1c_{mv}T_1+n_2c_{mv}T_2=n_{tot}c_{mv}T_f$

Oh yes from the conservation of internal energy ,thank you
But can't it be done by the way I have approached it??

 

[stockzahn]

Oh yes from the conservation of internal energy ,thank you
But can't it be done by the way I have approached it??

I'd say, if you are aware that in that specific case $p\left(V_1+V_2\right)=p_1V_1+p_2V_2$, but of course there could be other ways, I don't think of. A good chance to find an answer to that question is by asking @Chestermiller.

 

[Chestermiller]

You can't do it the way you said. The correct way to do this is $$n_1CT_1+n_2CT_2=(n_1+n_2)CT_f$$ with $$n_1=\frac{P_1V_1}{RT_1}$$ and $$n_2=\frac{P_2V_2}{RT_2}$$So, combining these equations, you get answer (c).

 

[Apashanka]

You can't do it the way you said. The correct way to do this is $$n_1CT_1+n_2CT_2=(n_1+n_2)CT_f$$ with $$n_1=\frac{P_1V_1}{RT_1}$$ and $$n_2=\frac{P_2V_2}{RT_2}$$So, combining these equations, you get answer (c).

Sir will you please explain why is the way I approached is not correct??

 

[Chestermiller]

FhhGffyuu

Sir will you please explain why it is ??

The first equation is just application of the first law of thermodynamics. The second and third equations are just application of the ideal gas law to determine the number of moles of gas in each chamber initially. Combining the three equations allows you to solve for the final temperature.

 

[Apashanka]

The first equation is just application of the first law of thermodynamics. The second and third equations are just application of the ideal gas law to determine the number of moles of gas in each chamber initially. Combining the three equations allows you to solve for the final temperature.

Yes sir actually I want to know ,why is the approach I made to solve this problem is incorrect??

 

[Chestermiller]

The final pressure does not require that p=p1 and p=p2. Your own equations already show that the final pressure is
$$P_f=\frac{(P_1V_1+P_2V_2)}{(V_1+V_2)}$$This value is somewhere between P1 and P2, and is equal to neither of them.

 

[Apashanka]

The final pressure does not require that p=p1 and p=p2. Your own equations already show that the final pressure is
$$P_f=\frac{(P_1V_1+P_2V_2)}{(V_1+V_2)}$$This value is somewhere between P1 and P2, and is equal to neither of them.

Yes sir I got it now