Boyle's law-effect of pressure at constant temperature imagine that you performed this experiment as described in your manual. in doing so, your first pressure reading was 660 mmhg, and your second presure reading was 780 mmig your first volume was 474 ml what would the percent error be if your measured second volume was 38.2 ml ? 4.76 x incorrect, one attempt remaining

Answer : The percentage error will be, 4.74%

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

or,

where,

= first pressure = 660 mmHg

= second pressure = 780 mmHg

= first volume = 47.4 mL

Note : (The value of first volume might be 47.4 mL because the difference between the value of first and second volume can not be more.)

= second volume = ?

Now put all the given values in the above equation, we get:

Thus, the value of second volume is, 40.1 mL

Now we have to calculate the percent error.

To calculate the percentage error, we use the equation:

We are given:

Experimental value of volume = 38.2 mL

Accepted value of volume = 40.1 mL

Putting values in above equation, we get:

Hence, the percentage error will be, 4.74%

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