A large rectangular vessel is divided by a partition into two compartments of id
ID: 1062824 • Letter: A
Question
A large rectangular vessel is divided by a partition into two compartments of identical shape, size, and temperature. One mole of fluorine gas is placed in the left compartment and one mole of nitrogen gas is placed in the right compartment. You may assume that both gases behave ideally.
(a) Which compartment has the greater pressure?
The left compartment has the greater pressure.
The right compartment has the greater pressure.
Both compartments have the same pressure.
(b) Which compartment has the greater density?
The left compartment has the greater density.
The right compartment has the greater density.
Both compartments have the same density.
(c) If the partition separating the two compartments is removed and the system is allowed to come to equilibrium, how will the pressure of each compartment compare with its original pressure?
The pressure in the left compartment will decrease, and the pressure in the right compartment will increase.
The pressure in the left compartment will increase, and the pressure in the right compartment will decrease.
The pressure in both compartments will remain unchanged.
(d) If the partition separating the two compartments is removed and the system is allowed to come to equilibrium, how will the density of each compartment compare with its original density?
The density in the left compartment will be lower and the density in the right compartment will be higher than the original density.
The density in the left compartment will be higher and the density in the right compartment will be lower than the original density.
The densities in both compartments will remain unchanged.
Explanation / Answer
a) Both compartments have the same pressure.
P*V =n*R*T
=> P depends only on V, T &n and not on Molecular weight or specific molecule.
b) The left compartment has the greater density.
P*M =d*R*T
=> d= P*M/(R*T)
where M= molecular weight
Therefore, density of Flourine is higher than N2
c) The pressure in both compartments will remain unchanged.
P1*V1/T1 =P2*V2/T2 (As n1= n2 =1mole)
Pressure & Temperature and volume are same
d) deq =0.5*d1 + 0.5*d2
The density in the left compartment will be lower and the density in the right compartment will be higher than the original density.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.