Question

Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy pT=12(pATA+pBTB)
when equilibrium is achieved.
Figure

Answer 1

PA' + PB'  = P

Explanation:

After connection = PA' →Partial pressure of A

PB'→ Partial pressure of B

PA' * 2V / T = PA * V/ TA

or PA' / T = PA/ 2TA -------(1)

Similarly PB' / T = PB/ 2TB -------(2)

Adding 1 and 2, we get

PA' / T  + PB' / T  = PA/ 2TA  + PB/ 2TB  

= 1/2 ( PA/TA + PB/TB)

P/T = = 1/2 ( PA/TA + PB/TB)  

Therefore PA' + PB'  = P

Answer 2

The temperature T and pressure P of vessel containing ideal gases satisfy  $\right)\\&\frac{P}{T}=\frac{1}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\end{aligned}$

Explanation:

Let P'A and P'B, respectively, be the partial gas pressure in chambers A and B.

Applying gas state equation A , we get

$\begin{aligned}\frac{P_{A} V}{T_{A}} &=\frac{P_{A}^{\prime} 2 V}{T} \\\frac{P_{A}}{T_{A}} &=\frac{2 P_{A}^{\prime}}{T}\end{aligned}$

$P_{A}^{\prime}=\frac{P_{A} T}{2 T_{A}}$

Similarly for Gas B

$P_{B}^{\prime}=\frac{P_{B} T}{2 T_{B}}$

Total pressure represents the sum of the partial pressure. It is administered by

$\begin{aligned}P &=P_{A}^{\prime}+P_{B}^{\prime} \\P &=\frac{P_{A} T}{2 T_{A}}+\frac{P_{B} T}{2 T_{B}}\end{aligned}$

$\begin{aligned}&P=\frac{T}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\\&\frac{P}{T}=\frac{1}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\end{aligned}$

The pressure P and temperature T of two vessels with rigid walls containing ideal gases satisfy $\right)\\&\frac{P}{T}=\frac{1}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\end{aligned}$ Hence the given condition has been satisfied.