A steel cylinder of oxygen is being store in a room at 25.0 °C under a pressure of 1250 atm. What pressure would be exerted by the gas if the container were transported across the desert at 40.0°C?

a. 1750 atm

b. 15700 atm

c. 1980 atm

d.1310 atm

P₂ = 1312.9 atm

Explanation:

Given data:

Initial temperature = 25°C

Initial pressure = 1250 atm

Final temperature = 40°C

Final pressure = ?

Solution:

Initial temperature = 25°C (25+273.15= 298.15 K)

Final temperature = 40°C (40+273.15 = 313.15 k)

According to Gay-Lussac Law,

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

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1250  atm /298.15 K = P₂/ 313.15 k

P₂ = 1250  atm × 313.15 k  / 298.15 K

P₂ = 391437.5 atm. K /298.15 K

P₂ = 1312.9 atm

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

the answer is c i took the test