A gas at 65degree c occupies 4.22l. What will be the volume of the gas at a temperature of 36.9degree c assuming constant pressure

Physics

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Step-by-step solution

4 steps
  1. Convert temperatures from Celsius to Kelvin.

    To convert Celsius to Kelvin, add 273.15.

    T1=65 ∘C+273.15=338.15 KT_1 = 65 \, ^\circ\text{C} + 273.15 = 338.15 \, \text{K} T2=36.9 ∘C+273.15=310.05 KT_2 = 36.9 \, ^\circ\text{C} + 273.15 = 310.05 \, \text{K}
  2. Identify the given values.

    Initial Volume (V1V_1) = 4.22 L4.22 \, \text{L}
    Initial Temperature (T1T_1) = 338.15 K338.15 \, \text{K}
    Final Temperature (T2T_2) = 310.05 K310.05 \, \text{K}
    Final Volume (V2V_2) = ?

  3. Apply Charles's Law.

    Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature:

    V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}

    To find V2V_2, rearrange the formula:

    V2=V1T2T1V_2 = \frac{V_1 T_2}{T_1}
  4. Substitute the values and calculate V2V_2.

    V2=(4.22 L)×(310.05 K)338.15 KV_2 = \frac{(4.22 \, \text{L}) \times (310.05 \, \text{K})}{338.15 \, \text{K}} V2=1308.411 L⋅K338.15 KV_2 = \frac{1308.411 \, \text{L} \cdot \text{K}}{338.15 \, \text{K}} V2≈3.869 LV_2 \approx 3.869 \, \text{L}

    The volume of the gas at 36.9 ∘C36.9 \, ^\circ\text{C} will be approximately 3.87 L.

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