Aballoon that had a volume of 3.50 l at 25.0°c is placed in a hot room at 40.0°c. if the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?

3.676 L.

Explanation:

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

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

(V₁T₂) = (V₂T₁)

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

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