Question

The apparatus shown consists of three temperature jacketed 1.000 L bulbs connected by stopcocks.Bulb A contains a mixture of H2O(g), CO2(g) and N2(g) at 25 C and a total pressure of 564 mm Hg. BulbB is empty and is held at a temperature of -70o C. Bulb C is also empty and is held at a temperatureof -190o C. The stopcocks are closed, and the volume of the lines connecting the bulbs is zero. CO2sublimes at -78o C, and N2 boils at -196 C.Q). The stopcock between A and B is opened and the system is allowed to come to equilibrium. Thepressure in A and B is now 219 mm Hg. Select correct alternate:(A) A contains CO2(g) and N2(g) and B contains CO2(g), N2(g) and H2O(s)(B) A contains CO2(g) and B contains N2(g) and H2O(l)(C) A contains CO2(g), N2(g) and H2O(s) and B contains CO2(g) and N2(g)(D) A contains H2O(g) and B contains H2O(g), N2(g) and CO2(g)Q)How many moles of H2O are in the system?(A) 0.026 mol (B) 0.0013 mol(C) 0.013 mol (D) 0.13 mol

​Both stopcocks are opened and the system is again allowed to come to equilibrium. The
pressure throughout the system is 33.5 mmHg. Select correct alternate:
(A) Each bulb contains N2(g), H2O(s) and CO2(g)
(B) Each bulb contains N2(g), H2O(g) and CO2(g)
(C) Each bulb contains N2(g), H2O(g) and CO2(s)
(D) A contains N2(g), B contains N2(g) and H2O(s); C contains N2(g) and CO2(s)

Q). How many moles of N2 are in the system?
(A) 0.022 mol (B) 0.011 mol
(C) 0.018 mol (D) 0.036 mol

Q). How many moles of
CO2 are in the system?
(A) 0.022 mol (B) 0.011 mol
(C) 0.018 mol (D) 0.036 mol​

Answer 1

"i) (A) $A contains CO_2(g) and N_2(g) and B contains CO_2(g), N_2(g) and H_2O(s)$

ii) (B) 0.0013 mol

Initial moles of gas = $n = PV/RT$

$=\frac{(564mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(298K)}$

Initial moles of gas = 0.3035 mol

Mole of gas in Bulb $A = n =PV/RT$

$=\frac{(219mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(298K)}=0.01178 mol$

Mole of gas in Bulb B = $n =PV/RT$

$=\frac{(219mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(203K)}=0.01729 mol$

$n_{H_2O}=n_{initial}-n_A-n_B=0.03035-0.01178-0.01729=0.00128mol=0.0013mol$

iii) (D) $A contains N_2(g), B contains N_2(g) and H_2O(s); C contains N_2(g) and CO_2(s)$

iv) (B) 0.011 mol

$n_A=PV/RT=\frac{(33.5mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(298K)}=0.001803 mol$

$n_B=PV/RT=\frac{(33.5mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(203K)}=0.002646 mol$

$n_C=PV/RT=\frac{(33.5mmHg\times\frac{1.00atm}{760mm Hg})(1.000L)}{(0.08206)(83K)}=0.006472 mol$

$N_2=n_A+n_B+n_C=0.001803+0.002646+0.006472=0.01092=0.011mol of N_2v) Irrelevant$"