Properties 38. A cylinder of compressed gas is labeled Compos ce N2.\" The 44. (

Question

Properties 38. A cylinder of compressed gas is labeled Compos ce N2." The 44. (mole %): 4.5% HS, 3.0% CO2, balance N,position pressure gauge attached to the cylinder reads 46 atn and Calculate the partial pressure of each gas, in atmo- spheres, in the cylinder 45. ure 46. 39) A halothane-oxygen mixture (GHBrCIF3 + 02) can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: P (halothane) - 170 mm Hg and P (02) = 570 mm Hg. (a) What is the ratio of the number of moles of halo- te Hg at obile thane to the number of moles of O2? (b) If the tank contains 160 g of O2, what mass of Diffus (See S 47 G,HBrCIF, is present? e the 40. A collapsed balloon is filled with He to a volume of 12.5 L at a pressure of 1.00 atm. Oxygen, O2, is then added so that the final volume of the balloon is 26 L with a total pressure of 1.00 atm. The temperature, which remains constant throughout, is 21.5 °C (a) What mass of He does the balloon contain? (b) What is the final partial pressure of He in the sure of O2 and 2O(g) in 02 75-L lete 22 C, Kinetic-Molecular Theory 48. balloon? (c) What is the partial pressure of O2 in the balloon? (d) What is the mole fraction of each gas? 49 (See Section 10-6 and Example 10.11.) wing 41.) You have two flasks of equal volume. Flask A 50. H2 at 0 C and 1 atm pressure. Flask B contains gas at 25 °C and 2 atm pressure. Compare t gases with respect to each of the following (a) average kinetic energy per molecule (b) root mean square speed (c) number of molecules (d) mass of gas 50-L be in the g of Nonid us 42. Equal masses of

Explanation / Answer

39) a) PA = nA(RT/V)    &     PB = nB(RT/V)

Let, A be Halothane and B be Oxygen.

So, PA/PB = nA/nB       (Temparetaure is contant)

Therefore, nA/nB = 170 mm Hg / 570 mm Hg

So, nA/nB = 17 / 57          (Ratio)

b)    Given, 160 g of O2

So, Moles of O2 = 160 / 32 = 5

Now, nA / nB = 17 / 57

So, nA / 5 = 17 / 57 ,

So, nA = 1.491 moles

Now, Halothane molecular weight is 197 g/mol

So, 1.491 moles of Halothane = 1.491 * 197 g/mol = 293.727 g

41)

a) KE = 3/2 kBT

TA = 0o C = 273 K and TB = 25o C = 273 + 25 = 298 K

So, KEA / KEB = TB / TA = 298 / 273 = 1.09

b) Vrms = ?(3RT / M)

Now, MA = 2 g/mol and MB = 44 g/mol

So, VA / VB = ?( MB * TA ) / ?( MA * TB ) = 4.48

c) PA = nA(RT/V)    &     PB = nB(RT/V)

So, PA/PB = (nA * TA ) / (nB * TB )

Therefore, nA/nB = (PA * TB ) / (PB * TA ) = 0.54

d) MassA / MassB = (nA * MA )/ (nB * MB ) = 0.0248             Where, MA = 2 g/mol & MB = 44 g/mol

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.