1- (A) A 5.10 L sample of neon at 10.11 atm is added to a 13.0 L cylinder that c
ID: 491334 • Letter: 1
Question
1- (A) A 5.10 L sample of neon at 10.11 atm is added to a 13.0 L cylinder that contains argon. If the pressure in the cylinder is 7.92 atm after the neon is added, what was the original pressure (in atm) of argon in the cylinder?
(B)Calculate the total pressure, in atm, in a 5 L flask that contains 6.42 g of Ne and 1.55 g of Ar. The temperature of the gases is 32 oC.
(C)Calculate vrms, the root mean square velocity, in m/s of SO2 molecules at 274 oC.
(D)An effusion container is filled with 7 L of an unknown gas. It takes 132 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 335 s for 7.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas.
(E)Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 28.0 mL. The first bulb has a volume of 83.0 mL and contains 5.10 atm of argon, the second bulb has a volume of 250 mL and contains 1.09 atm of neon, and the third bulb has a volume of 33.0 mL and contains 4.04 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm?
(F)Use the van der Waals equation to calculate the pressure, in atm, of 45.61 mol of hydrogen at 443 oC in a 2.45 L container.
Explanation / Answer
Hi
A) P1= 10.11 atm V1= 5.10 L,
n1= RT/P1V1
P2= ?, V2= 13 L,
n2= RT/P2V2,
P3= 7.92 atm, V3= 13 L
n3= n1 + n2
(P3V3/RT) = (P1V1/RT) + (P2V2/RT)
P3V3= P1V1+ P2V2
7.92*13= 10.11*5.10+ P2*13
P2= 98.99 atm
B) P= ?, V= 5 L, moles of neon (n1)= 6.42/20= 0.321 gmole
moles of argon (n2)= 1.55/40 0.03875 gmole
T= 32 oC= 305 K
R= 0.08205 atm-L/mole K
PV= (n1 + n2)RT
P*5= (0.321+0.03875)*0.08205*305
P= 1.80 atm
Thank you
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.