Calculate the pressure of 2.78 g of Ne in a 5.00 L flask at -10°C

0.60 atm

Explanation:

Step 1.

Data obtained from the question.

This includes the following:

Mass of Ne = 2.78g

Volume (V) = 5L

Temperature (T) = –10°C

Pressure (P) =?

Step 2:

Determination of the number of mole of Ne.

The number of mole of Ne can be obtained as follow:

Mass of Ne = 2.78g

Molar Mass of Ne = 20g/mol

Number of mole Ne =?

Number of mole = Mass/Molar Mass

Number of mole of Ne = 2.78/20

Number of mole of Ne = 0.139 mol

Step 3:

Conversion of celsius temperature to Kelvin temperature. This can be achieved by as shown below:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = –10°C

Temperature (Kelvin) = –10°C + 273

Temperature (Kelvin) = 263K

Step 4:

Determination of the pressure. The pressure can be obtained by using the ideal gas equation as follow:

PV = nRT

Volume (V) = 5L

Temperature (T) = 263K

Number of mole (n) of Ne = 0.139mol

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =?

PV = nRT

P x 5 = 0.139 x 0.082 x 263

Divide both side by 5

P = (0.139 x 0.082 x 263) /5

P = 0.60 atm

Therefore the pressure is 0.60 atm

catya that's not even a b or c

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