A 1.2 L weather balloon on the

ground has a temperature of 25°C

and is at atmospheric pressure (1.0

atm). When it rises to an elevation

where the pressure is 0.74 atm,

then the new volume is 1.8 L. What

is the temperature (in °C) of the air

at this elevation?

330.95K

Explanation:

V₁ = 1.2L

T₁ = 25°C = (25 + 273.15)K = 298.15K

P₁ = 1.0 atm

P₂ = 0.74 atm

V₂ = 1.8L

T₂ =?

From combined gas equation,

(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂

Solve for T₂

T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)

T₂ = (0.74 * 1.8 * 298.15) / (1.0 * 1.2)

T₂ = 397.1358 / 1.2

T₂ = 330.9465K

T₂ = 330.95K or T₂ = (330.95 - 273.15)°C = 57.8°C

the answer is d.) potassium

the answers are

c and c