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Question: A glass bulb of volume \(400c{m^3}\) is connected to another of volume \(200c{m^3}\) by means of a t...

A glass bulb of volume 400cm3400c{m^3} is connected to another of volume 200cm3200c{m^3} by means of a tube of negligible volume. The bulbs contain dry air and both are at a common temperature and pressure of 20∘C20^\circ C and 1.00 atm respectively. The larger bulb is immersed in steam at and the smaller in melting ice at 0∘C0^\circ C. Find the final common pressure.
(A) 6.13 atm
(B) 7.13 atm
(C) 2.13 atm
(D) 1.13 atm

Explanation

Solution

The amount of fluid inside the chambers is constant i.e. the number of moles of the dry air is constant. We can use the ideal gas equation to find the final pressure,

PV=nRT nR=PVT=const  PV = nRT \\\ nR = \dfrac{{PV}}{T} = const \\\

Complete step by step answer:
There is an initial condition given which states that at a common temperature and pressure there is dry air in two glass bulbs. When the temperature is changed individually, there will be an instability. After some time at a common pressure the system will get stable, number of moles being the same. The ideal gas equation is given by,
PV=nRTPV = nRT
Where, P = pressure
V= volume
n= number of moles of gas
R= gas constant
T=temperature
On rearranging we have,

nR=PVT=const ⇒P1V1T1=P2V2T2=const  nR = \dfrac{{PV}}{T} = const \\\ \Rightarrow \dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} = const \\\

For a more general case involving multiple domains such as the present case, we have,
∑iP1,iV1,iT1,i=∑iP2,iV2,iT2,i\sum\limits_i {\dfrac{{{P_{1,i}}{V_{1,i}}}}{{{T_{1,i}}}} = \sum\limits_i {\dfrac{{{P_{2,i}}{V_{2,i}}}}{{{T_{2,i}}}}} }
Substituting the values, we get,

⇒1(400)293+1(200)293=P(400283+200273) P=(400)293+(200)293400283+200273atm P=1.13atm  \Rightarrow \dfrac{{1(400)}}{{293}} + \dfrac{{1(200)}}{{293}} = P(\dfrac{{400}}{{283}} + \dfrac{{200}}{{273}}) \\\ P = \dfrac{{\dfrac{{(400)}}{{293}} + \dfrac{{(200)}}{{293}}}}{{\dfrac{{400}}{{283}} + \dfrac{{200}}{{273}}}}atm \\\ P = 1.13atm \\\

The final common pressure is 1.13 atm.
The correct answer is option D.

Note: In solving gas equation problems, be careful with the units. Here, a constant known as gas constant is involved which has different values for different sets of units. Therefore, while substituting the values, ensure that a given parameter has the same unit as that in the gas constant. Change the other parameter units accordingly.
Here we have not converted the units as the gas constant gets cancelled and is not involved in final calculation. Also, the units of other parameters get eventually cancelled and we are left with pressure (atm).