Question

Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at ${27^\circ }C$ and contain $0.7$ mol of ${H_2}$ at $0.5{\text{ atm}}{\text{.}}$. One of the flasks is then immersed into a bath kept at ${127^\circ }C$, while the other remains at ${27^\circ }C$. The final pressure in each flask is:
(A) Final pressure in the apparatus is $0.5714{\text{ atm}}$
(B) Number of moles in one of the flasks are $0.4$ and in other are $0.3$
(C) Final pressure is $3.7281{\text{ atm}}$
(D) Both (A) and (B) are correct

Answer 1

Hint:These type of numerical are based on the ideal gas equation which are present in the form of $PV = nRT$ in which $P$ is the pressure, $V$ is volume, $n$ is the amount and $R$ is universal gas constant and $T$ is temperature of an ideal gas. Ideal gas equation is also used for calculating the pressure, volume of gases which is consumed or produced value in the two flasks which are given in these questions.

Complete step by step answer:The given information in these question is that there are two flasks of equal volume at initial temperature if ${27^\circ }C$ , therefore initial pressure is $0.5{\text{ atm}}{\text{.}}$ and it contains $0.7{\text{ moles}}$ of ${H_2}$ , we calculate the final pressure in each flask.
So, firstly there are two flasks which are marked as $a$ and $b$, the first flask is $a$ and the second is $b$.
The first flask $'a'$ be immersed in a both at temperature, ${T_a}$ is ${127^\circ }$ C which are converted into kelvin.
So, ${T_a} = 127 + 273.15K\left( {{O^\circ }C = 273.15K} \right)$
\[{T_a} = 400.15K\]
The second flask $'b'$ have the same as initial temperature where \[{T_b}\] is ${27^\circ }$ C which are convert into kelvin
So, \[{T_b} = {27^\circ }C + 273.15K\]
\[{T_b} = 300.15K\]
In which number of moles of two flask which is $a$ and $b$ are marked as \[{n_a}\] and \[{n_b}\]
Then, the final pressure and volume of two both flasks are equal and same value in which means that \[P = {P_a} = {P_b}\], \[V = {V_a} = {V_b}\]
Ideal gas law for the combined system are using initially that is,
\[PV = nRT\]…….. (i)
Put the value of $P$ is \[0.5\] , $V = 2V$ because the two both flask are same volume and the number of moles
$n$ is $0.7$ and the temperature is \[300.15K\] in the equation (i)
 \[0.5 \times 2V = 0.7 \times R \times 300.15\]
\[2V = \dfrac{7}{{\not{10}}} \times \dfrac{{\not{10}}}{5} \times 300.15 \times R\]
\[V = \dfrac{7}{{5 \times 2}} \times 300.15 \times R\]
\[V = \dfrac{{7 \times 300.15 \times R}}{{10}}\]
\[V = 210.105R\] …….. (ii)
The ideal gas equation for flask first (a):
\[{P_a} \times {V_a} = {n_a} \times R \times {T_a}\]
Where, \[{P_a} = P\]and \[{V_a} = V\] put in these,
\[P \times V = {n_a} \times R \times {T_a}\]
 Put the value of $V$ and ${T_a}$ in the equation
\[P \times 210.105R = {n_a} \times R \times 400.15\] in which the value of $V$ from equation (ii) put,
\[{n_a} = 0.525P\] ……… (iii)
Then, the ideal gas equation for flask second (b)
\[{P_b} \times {V_b} = {n_b} \times R \times {T_b}\]
Where, \[{P_b} = P\] and \[{V_b} = V\] put in these
\[P \times V = {n_b} \times R \times {T_b}\]
From equation (ii) value of $V$ is put and \[{T_b}\] also
\[P \times 210.105R = {n_b} \times R \times 300.15\]
\[{n_b} = \dfrac{{P \times 210.105R}}{{R \times 300.15}}\]
\[{n_b} = 0.7P\] ………. (iv)
So, here as we told that
Total number of mole initially is equal to sum of number of moles of flask a and number of moles of flask b
Total number of moles initially \[ = {n_a} + {n_b}\]
Put the value of \[{n_a}\]and \[{n_b}\]from equation (iii) and (iv) equation
\[0.7 = 0.525P + 0.7P\]
\[0.7 = 1.225P\]
\[P = \dfrac{{0.7}}{{1.225}} = 0.5714{\text{ atm}}\]
So, the final pressure of each flask is \[0.5714{\text{ atm}}\].
Hence option A is the correct answer.


Note:An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.