A sample of neon gas occupies a volume of 27L 26 degrees * C . What is the final temperature of the sample if the volume is decreased to 59L? help please

380 °C

Explanation:

The following data were obtained from the question:

Initial volume (V₁) = 27 L

Initial temperature (T₁) = 26 °C

Final volume (V₂) = 59 L

Final temperature (T₂) =?

Next, we shall convert 26 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 26 °C

Initial temperature (T₁) = 26 °C + 273

Initial temperature (T₁) = 299 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 27 L

Initial temperature (T₁) = 299 K

Final volume (V₂) = 59 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ /T₂

27 / 299 = 59 / T₂

Cross multiply

27 × T₂ = 299 × 59

27 × T₂ = 17641

Divide both side by 27

T₂ = 17641 / 27

T₂ = 653 K

Finally, we shall convert 653 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 653 K

T₂ = 653 – 273

T₂ = 380 °C

Thus, the final temperature of the gas is 380 °C

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).