12. A helium-filled weather balloon has a volume of 2.4 x 102 L at 99 kPa pressure and a
temperature of 0°C. What is the mass of the helium in the balloon? (Hint: Helium is a
diatomic gas.)

The answer to your question is 8.74 g of He

Explanation:

Data

V = 2.4 x 10² L

P = 99 kPa

T = 0°C

mass = ?

Process

1.- Convert kPa to atm

P = 99 kPa = 99000 Pa

                   1 atm 101325 Pa

                    x         99000 Pa

                   x = (99000 x 1) / 101325

                   x = 0.977 atm

2.- Convert temperature to °K

°K = 273 + 0

°K = 273

3.- Substitution

      PV = nRT

- Solve for n

      n = PV / RT

      n = (0.977)(2.4 x 10²) / (0.082)(273)

      n = 24.48 / 22.386

      n = 1.093 moles

4.- Calculate the grams of He

          8 g 1 mol

           x     1.093 moles

           x = (1.093 x 8) / 1

           x = 8.74 g                      

The mass of the helium gas is 42 grams

Explanation:

Step 1: Data given

Volume of the balloon = 2.4 *10^2 L = 240 L

Temperature = 0°C = 273 K

Pressure = 99 kPa = 0.977054 atm

Molar mass He2 = 4.0 g/mol

Step 2: Calculate moles He

p*V = n*R*T

⇒with p = the pressure = 0.977054 atm

⇒with V = the volume = 240 L

⇒with n = the number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 273 K

n = (p*V) / (R*T)

n = (0.977054 * 240 ) / (0.08206 * 273)

n = 10.467 moles

Step 3: Calculate mass of Helium

Mass He = Moles He * molar mass He

Mass He = 10.467 moles * 4.0 g/mol

Mass He = 42 grams

The mass of the helium gas is 42 grams

NOTE: The noble gases (helium, neon, argon, krypton, xenon, and radon) are gases at STP and are monatomic

bcde is the incorrect answer it is a

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