A sample of ammonia gas has a volume of 3213 mL at 11°C and a pressure of 822 torr. What will the volume of the gas be in liters if the moles of gas and the temperature do not change but the pressure changes to 2.33 atm?

We are given:

v1 = 3.2 L                   v2 = x L

p1 = 822 torr OR 1.08 atm        p2 = 2.33 atm

t1 = t2 = 284 k

From the Ideal gas equation:

PV = nRT

Since the number of moles (n), universal gas constant (R) and Temperature (T) are constant

PV = k      (where k is a constant)

it can also be written as:

P1V1 = k (1)

Similarly,

P2V2 = k (where k is the same constant as before)(2)

Solving for V2:

From (1) and (2):

P1V1 = P2V2

Replacing the variables

1.08 * 3.2 = 2.33 * x

x = (1.08 * 3.2)/2.33

x = 1.5 L (approx)

Therefore, the final volume of the solution will be 1.5L

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole