Question

1. A 4.00 L sample of air at 35 degrees C expands to 5.50 L when heated. What is

Answer 1

Answer:

150.5 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 35 °C

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

Next, we shall convert 35 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 35 °C

Initial temperature (T₁) = 35 °C + 273

Initial temperature (T₁) = 308 K

Next, we shall determine the final (i.e the new) temperature of air. This can be obtained as follow:

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 308 K

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

4 / 308 = 5.5 / T₂

Cross multiply

4 × T₂ = 308 × 5.5

4 × T₂ = 1694

Divide both side by 4

T₂ = 1694 / 4

T₂ = 423.5 K

Finally, we shall convert 423.5 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 423.5 K

T₂ = 423.5 – 273

T₂ = 150.5 °C

Thus, the new temperature of the air is 150.5 °C