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17 January, 10:01

425 mL of neon gas at - 12°C and 788 mmHg

expands into a 2.40 L container while the

pressure decreases to 577 mmHg. What is the

temperature of the gas in °C?

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Answers (1)
  1. 17 January, 11:22
    0
    806.23 °C

    Explanation:

    The question requires we calculate the new temperature of the gas in °C.

    We are given;

    Initial volume of neon gas, V₁ = 425 mL or 0.425 L Initial temperature of neon gas, T₁ = - 12°C

    When doing questions on gas law, we always use temperature in Kelvin.

    To convert temperature from °C to Kelvin we always use;

    K = °C + 273

    Therefore, Initial temperature, T₁ = 261 K

    Initial Pressure, P₁ = 788 mmHg New pressure of the gas, P₂ = 577 mmHg New volume of the gas, V₂ = 2.40 L

    To calculate the new temperature, we are going to use the combined gas equation.

    According to the combined gas law, P₁V₁/T₁ = P₂V₂/T₂ Rearranging the equation;

    T₂ = P₂V₂T₁ / P₁V₁

    Therefore;

    T₂ = (577 mmHg * 2.4 L * 261 K) : (788 mmHg * 0.425 L)

    = 1079.226 K

    = 1079.23 K

    But, °C = K - 273

    = 1079.23 K - 273

    Therefore, T₂ = 806.23 °C

    Therefore, the new temperature is 806.23°C
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