Which of the following will cause the volume of an ideal gas to triple in value? a) raising the absolute temperature by a factor of 3 while increasing the pressure by a factor of 3.b) lowering the absolute temperature by a factor of 3 while increasing the pressure by a factor of 3.c) lowering the pressure by a factor of 3 while the temperature stays constant. d) lowering the absolute temperature by a factor of 3 at constant pressure. e) raising the temperature from 25 c to 75 c at constant pressure.

C) C) Lowering the pressure by a factor of 3 while the temperature stays constant.

Explanation:

P V = nRT ⇒  V = nRT/P

Volume is directly proportional to  n, T and inversely proportional to P

a) Vfinal  = n R (3TI ) /3Pi = nRTi/Pi = Vi                                  FALSE

b) Vinal = n R (Ti/3) /(3Pi) = nRTi/Pi = (1/9) Vinitial                  FALSE

c)  Vfinal = n R(Ti/3) / Pi = nRTi/Pi = Vi                                    TRUE  

d) Vfinal = nR(Tinitial/3P = 1/3 Vi                                             FALSE

e) False we do not use ªC in the ideal gas law, we use K    FALSE

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years