3. methane (ch4) behaves as an ideal gas under standard temperature and pressure conditions. what volume is occupied by 1 kg of methane at a temperature of 35 °c and a pressure of 0.9 atm?

1711.11L

Explanation:

Given parameters:

Mass of methane = 1kg = 1000

Temperature of gas = 35°C

Pressure of the gas = 0.9atm

Converting to the temperature to kelvin, we have:

               35+ 273 = 308k

Solution

From the given parameters, we can solve this problem using the ideal gas equation which is combination of the gas laws. It is expressed below:

                                         PV = nRT              .......................(i)

Where P is the pressure of the gas

             V is the volume of the gas

             n is the number of moles

             R is the gas constant

             T is the temperature

Now, from the above parameters, we have two unknowns which are n and V.

To find the number of moles n, we use the expression below:

         number of moles =

        molar mass of methane = 12 + (4x1) = 16g/mol

       Number of moels of methane =   = 62.5mole

Now that we know the number of moles, we can make the unknown Volume the subject of the formula:

                     V =

                     V =   = 1711.11L

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

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