Consider a rigid, insulated box containing 0.400 mole of he(g) at 20.08c and 1.00 atm in one compartment and 0.600 mole of n2(g) at 100.08c and 2.00 atm in the other compartment. these compartments are connected by a partition that transmits heat. what is the final tempera-ture in the box at thermal equilibrium? [for he(g), cv 5 12.5 j k 21mol 21 ; for n2(g), cv 5 20.7 j k 21mol 21

The value of final temperature of the mixture of gases when the partition is removed = 341.08 K

Explanation:

For box 1

Mass = 0.4 mole

Temperature = 20.08 °c = 293.08 K

Pressure = 1 atm = 101.325 K pa

For box 2

Mass = 0.6 mole

Temperature = 373.08 K

Pressure = 2 atm = 202 K pa

When the partition removed, both gases mixes together. then the final mass

= +

= 0.4 + 0.6

= 1.0 mole

This is the final mass in the container.

Final temperature is given by the equation,

+ =

Put all the values in the above equation we get,

0.4 × 293.08 + 0.6 × 373.08 = 1  ×

= 341.08 K

This is the value of final temperature of the mixture of gases when the partition is removed.

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