At 27.0°C, the volume of a gas is 630 L. At the same pressure, its volume is 92,0 mL at a temperature of

Question

At 27.0°C, the volume of a gas is 630 L. At the same pressure, its volume is 92,0 mL at a temperature of

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Philomena 5 years 2021-08-14T23:03:05+00:00 1 Answers 16 views 0

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  1. Farah
    0
    2021-08-14T23:04:42+00:00
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    Answer:

    –272.96 °C

    Explanation:

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

    Initial temperature (T₁) = 27.0 °C

    Initial volume (V₁) = 630 L.

    Final volume (V₂) = 92.0 mL

    Final temperature (T₂) =?

    Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:

    T(K) = T(°C) + 273

    Initial temperature (T₁) = 27.0 °C

    Initial temperature (T₁) = 27.0 °C + 273

    Initial temperature (T₁) = 300 K

    Next, we shall convert 92.0 mL to L. This can be obtained as follow:

    1000 mL = 1 L

    Therefore,

    92 mL = 92 mL × 1 L / 1000 mL

    92 mL = 0.092 L

    Next, we shall determine the final temperature.

    Initial temperature (T₁) = 300 K

    Initial volume (V₁) = 630 L.

    Final volume (V₂) = 0.092 L

    Final temperature (T₂) =?

    V₁ / T₁ = V₂ / T₂

    630 / 300 = 0.092 / T₂

    2.1 = 0.092 / T₂

    Cross multiply

    2.1 × T₂ = 0.092

    Divide both side by 2.1

    T₂ = 0.092 / 2.1

    T₂ = 0.04 K

    Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:

    T(°C) = T(K) – 273

    Final temperature (T₂) = 0.04 K

    Final temperature (T₂) = 0.04 – 273

    Final temperature (T₂) = –272.96 °C

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