A helium-filled weather balloon has a volume of 697 L at 22.9°C and 754 mmHg. It is released and rises to an altitude of 6.93 km, where the pressure is 377 mmHg and the temperature is –23.1°C. The volume of the balloon at this altitude is L.

The new volume at this altitude is 1177.3 L

Explanation:

Step 1: Data given

Volume of a helium filled balloon = 697 L

Temperature = 22.9 °C = 295.9 K

Pressure = 754 mmHg

The pressure reduces to 377 mmHg

The temperature decreases to -23.1 °C = 249.9 K

Step 2: Calculate the new volume at this altitude

P1 *V1 / T1  = P2*V2 / T2

⇒with P1 = the initial pressure of the gas = 754 mmHg

⇒with V1 = the initial volume in the balloon = 697 L

⇒with T1 = the initial volume = 295.9 K

⇒with P2 = the reduced pressure = 377 mmHg

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature = 249.9 K

(754 mmHg* 697 L ) / 295.9 K = (377 mmHg * V2) / 249.9 K

V2 =  1177.3 L

The new volume at this altitude is 1177.3 L

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