Answer:
1. 2.33 L.
2. 111.74 °C.
Explanation:
1. 3.0 L of a gas is at a temperature of 78°C. Find the volume of the gas at standard temperature.
- Charles’ Law states that at constant pressure, the volume of a given quantity of a gas varies directly with its temperature.
V ∝ T.
∴ V₁T₂ = V₂T₁.
V₁ = 3.0 L, T₁ = 78°C + 273 = 351 K.
Standard temperature = 0.0 °C.
V₂ = ??? L, T₂ = 0.0 °C + 273 = 273 K.
∴ V₂ = V₁T₂/T₁ = (3.0 L)(273 K)/(351 K) = 2.33 L.
2. At what temperature (in oC) would the volume of a gas be equal to 45.7 L, if the volume of that gas was 33.9 L at 12.4oC?
∵ V₁T₂ = V₂T₁.
V₁ = 33.9 L, T₁ = 12.4°C + 273 = 285.4 K.
V₂ = 45.7 L, T₂ = ??? K.
∴ T₂ = V₂T₁/V₁ = (45.7 L)(285.4 K)/(33.9 L) = 384.74 K.
∵ T(K) = T(°C) + 273.
∴ 384.74 K = T(°C) + 273.
∴ T(°C) = 384.74 K - 273 = 111.74 °C.
