10.1 and 10.2 Gas Characteristics; Pressure (10.13 Brown) How does a gas compare

Question

10.1 and 10.2 Gas Characteristics; Pressure (10.13 Brown) How does a gas compare with a liquid for each of the following properties a. 1. Density b. Compressibility Ability to mix with other substances of the same phase to form homogeneous mixtures d. c. Ability to conform to the shape of its container (10.22 Brown) Perform the following conversions a. b. c. d. 2. 0.912 atm to torr 0.685 bar to kilopascals 655 mm Hg to atmospheres 1.323x10% Pa to atmospheres Open end Open end Closed end (10.25 Brown) If the atmospheric pressure is 0.995 atm, what is the pressure of the enclosed gas in each of the three cases depicted in the drawing? Assume that the gray liquid is mercury 3. Gas Gas h-52 cm h=67cm h 10.3 cm 10.3 and 10.4 Gas Laws; Ideal Gas Equation (in) (iin) (10.28 Brown) A fixed quantity of gas at 21°C exhibits a pressure of 752 torr and occupies a volume of 5.12 L a. 4. Calculate the volume the gas will occupy if pressure is increased to 1.88 atm while the temperature is held constant Calculate the volume the gas will occupy if the temperature is increased to 175°C while the pressure is held constant. b. 5. (10.35 Brown) Complete the following table for an ideal gas 1.00 L 0.500 mol 2.00 atm 0.300 atm 650 torr 27°C 350 K 294 K 0.250 L mol 0.333 mol 585 ml 0.250 mol atm (10.38 Brown) A neon sign is made of glass tubing whose inside diameter is 2.5 cm and whose length is 5.5m. If the sign contains neon at a pressure of 1.78 torr at 35°C, how many grams of neon are in the sign? (Volume of cylinder ) 6.

Explanation / Answer

1a) In gases , particles are far from each other thus occupy large volume which results in lesser density than liquids in which particles are comparatively closely spaced. ( density is mass by volume)

b)gases are highly compressible due to presence of large interparticle spaces.

c) Gases mix with other gases to form homogeneous mixture whereas liquid mix with another liquid to form a mixture which may be homo or heterogeneous.

d) Gases due to large interparticle space and weak interparticle forces of attraction occupy the available space whereas liquids have definite volume they acquire the shape of a container in which they are placed.

2 a) 0.912 atm = 0.912*760 torr (1atm=760torr)

=693.1 torr

b)0.685bar = 0.685*100kpa =68.5kpa

c)655mm Hg = 655/760 atm =0.86atm

d)1.323*105Pa = 1.323 *105* 9.869 *10-6 =1.3atm

3) here we use formula

P= Po + h dg

Po= atmosperic pressure =0.995atm = 0.995*106dyne cm-2

d= density of mercury=13.6g/ml

g= 980cm/s2, h= difference in height of column

i) P=0.995*106 +52*980*13.6= 1.6*106 dyne cm-2

ii) P=995000+67*980*13.6=1.9*106 dyne cm-2

iii) as tube is closed Po =0  

So, P=hdg =10.3*13.6*980= 1.4* 105 dyne/cm2

4)a) as T is constant So using boyle law

P1V1 =P2 V2

752 *5.12 = 1.88*760 *V2

V2 = 2.7 L

b)At constant pressure

V1 T2 = V2 T1 ( T is in kelvin)

5.12* 448 = V2 *294

V2 =7.8L

5) using ideal gas equation in each case

PV =n R T

R =gas constant = 0.0821L atm /mol K

i) T =PV/nR =2*1/0.5*0.0821 = 48.78K

ii)n= PV/RT = 0.3* 0.25/0.082*300 = 0.003mol

iii)V= nRT/P =0.333*0.082*350/650/760= 1.9*10-5 L

iv)P =n RT/V = 0.250*0.082*294/0.585 = 10.3atm

6) V=r2h= 3.14 * 2.5/200 *2.5/200 *5.5 = 0.0027m3 =2.7L

PV=nRT

1.78/760 atm *2.7L = n 0.082atm L/mol/K* 308K

n= 0.00025mol

= 0.00025*40g =0.01g

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