A 2.00-L glass soda bottle filled only with air is tightly capped at 18 degree C

Question

A 2.00-L glass soda bottle filled only with air is tightly capped at 18 degree C and 718.0 mmHg. If the bottle is placed in water at 71 degree C, what is the pressure in the bottle? a. 182 mmHg b. 849 mmHg c. 2830 mmHg d. 607 mmHg e. 428 mmHg Which of the following statements are postulates of the kinetic-molecular theory of gases? Gas particles are in constant, random motion. 2. The distance between gas particles is large in comparison to their size. 3. The average kinetic energy of gas particles is proportional to the kelvin temperature. a. 1 only b. 2 only c. 3 only d. 1 and 2 e. 1, 2, and 3 Sodium azide decomposes rapidly to produce nitrogen gas. 2 NaN_3(s) rightarrow 2 Na(s) + 3 N_2(g) What mass of sodium azide will inflate a 60.0 L airbag for a car to a pressure of 1.50 atm at 32 degree C? (R = 0.08206 L atm/mol K) a. 2.40 g b. 67.2 g c. 156 g d. 234 g e. 351 g What is the pressure of a 49.4-L gas sample containing 1.04 mol of gas at 36.4 degree C? (R = 00821 L middot atm/K middot mol), 1 atm = 760 mmHg) a. 4.78 times 10^1 mmHg b 7.04 times 10^-4 mmHg c. 0.534 mmHg d. 4.06 times 10^2 mmHg e. 1.42 times 10^3 mmHg The volume of a sample of gas measured at 65.0 degree C and 1.00 atm pressure is 2.00 L. What must the final temperature be in order for the gas to have a final volume of 8.00 L at 1.00 atm pressure? A. 1079.0 degree C b. 260.0 degree C c.-13.0 degree C d. 16.3 degree C e. -188.5 degree C

Explanation / Answer

36)

PV=nRT

since volume, moles, and R are constant, then:

P1/T1=P2/T2

P1=718 mmHg; T1=291K
P2=?; T2=344K

(718 mmHg)/(291K)=P2/(344K)

solve for P2

P2=848 mmHg

Answer : B (849 mmHg)

37)

Answer: e (1,2 and 3)

38)
V= 60.0 L
P= 1.50 atm
T= 32°C 305.15 K
n= ?

THE IDEAL GAS LAW ;

P•V = n•R•T n=P•V/R•T

n=1.50 atm*60.0L/(0.0821 (L•atm/mol•K)*305.15K)
n=3.59 moles of N2

Using the Stoichiometry;
Moles of NaN2 = 2* 3.59 / 3 = 2.393 moles
Mass of NaN2 = moles * moleculat wt = 2.393 moles * 65 g/mol = 155.56 g
Mass of NaN2 = 156 g

Answer: C

39) what is the pressure of a 49.4 L gas sample containing 1.04 mol of gas at 36.4 oC

P = nRT/V = 1.04 (mol) * 0.0821 (L•atm/mol•K)* 309.55 (K)/49.4 (L) = 0.535 atm

P = 0.535 atm = 0.535 atm * 760 mmHg = 406 mmHg = 4.06 x 10^2 mmHg

Answer : D

40) the volume of a sample of gas measured at 65oC and 1 atm pressure is 2 L. what must the
final temperature be in order for the gas to have a final volume of 8 L at 1 atm pressure

P1V1/ T1 = P2V2/T2

(1 atm * 2L / 338.15K) = (1 atm * 8L/ T2)
T2 = 1352.6 K = 1079.45 oC = 1079 oC
Answer : A

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