A sample of gas has an initial volume of 30.8L AND an initial temperature of -67 degree Celcius. What will be the temperature of the gas if the volume is 21.0L?

–132.55 °C

Explanation:

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

Initial volume (V₁) = 30.8 L

Initial temperature (T₁) = –67 °C

Final volume (V₂) = 21.0 L

Final temperature (T₂) =?

Next, we shall convert –67 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = –67 °C

Initial temperature (T₁) = –67 °C + 273

Initial temperature (T₁) = 206 K

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

Initial volume (V₁) = 30.8 L

Initial temperature (T₁) = 206 K

Final volume (V₂) = 21.0 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

30.8 / 206 = 21 / T₂

Cross multiply

30.8 × T₂ = 206 × 21

30.8 × T₂ = 4326

Divide both side by 30.8

T₂ = 4326 / 30.8

T₂ = 140.45 K

Finally, we shall convert 140.45 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 140.45 K

T₂ = 140.45 K – 273

T₂ = –132.55 °C

Thus, the new temperature of the gas is –132.55 °C

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