7. A sample of argon gas has a volume of 0.43 mL at 24 C. At what temperature in Celsius
will it have a volume of 0.57 mL?

120.55 °C

Explanation:

Given data:

Initial volume of gas = 0.43 mL

Final volume of gas = 0.57 mL

Initial temperature = 24°C (24+273 = 297 k)

Final temperature = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁  

T₂ = 0.57 mL × 297 K / 0.43 mL

T₂ = 169.29 mL.K / 0.43 mL

T₂ = 393.7 K

Temperature in celsius:

393.7 K - 273.15 = 120.55 °C

hurricanes form when warm, moist air from the ocean surface quickly rises into the atmosphere and forms storm clouds and rain. when the clouds are formed, heat is released which warms the cool air in the upper regions of the atmosphere so that the cycle can continue. which two phases of the water cycle are repeated in this process?

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