Before it was launched, a helium-filled balloon had a pressure of 101 kpa at a temperature of 20°c. at an altitude of 15 000 m, the pressure had decreased to 11.7 kpa and the temperature had dropped to -56 °c. the volume of the balloon increased to 35. 4 m3.

what is the original volume of the ballon
6.9 m3
1.34 m3
2.56 m3
43.0 m3
5.54 m3

5.54 m³.

Explanation:

where, P is the pressure of the gas in atm,

V is the volume of the gas in L,

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K.

We have two different cases at different (P, V and T) while the number of moles of He and R is constants.

∴ (P₁V₁) / (T₁) = (P₂V₂) / (T₂).

We can use P in KPa and V in m³ that the conversion factor can be canceled by division, but we should use T in K because its conversion factor is additive value.

P₁ = 101.0 kPa, V₁ = ??? m³, and T₁ = (20 °C + 273) = 293.0 K.

P₂ = 11.7 kPa, V₂ = 35.4 m³, and T₂ = (-56 °C + 273) = 217 K.

∴ The initial volume (V₁) = (P₂V₂T₁) / (P₁T₂) = (11.7 kPa)(35.4 m³)(293.0 K) / (101.0 kPa)(217.0 K) = (121354.74) / (21917) = 5.537 m³ ≅ 5.54 m³.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.