A quantity of gas occupies a volume of 804cm³ at a temperature of 127degree c . At what temperature will the volume of the gas be 603cm³ assuming that there is no change in the pressure

Physics

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Step-by-step solution

5 steps
  1. Convert the initial temperature from Celsius to Kelvin.

    To convert Celsius to Kelvin, add 273.15.

    T1=127 ∘C+273.15=400.15 KT_1 = 127 \, ^\circ\text{C} + 273.15 = 400.15 \, \text{K}
  2. Identify the given values.

    Initial Volume (V1V_1) = 804 cm3804 \, \text{cm}^3
    Initial Temperature (T1T_1) = 400.15 K400.15 \, \text{K}
    Final Volume (V2V_2) = 603 cm3603 \, \text{cm}^3
    Final Temperature (T2T_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 T2T_2, rearrange the formula:

    T2=V2T1V1T_2 = \frac{V_2 T_1}{V_1}
  4. Substitute the values and calculate T2T_2.

    T2=(603 cm3)×(400.15 K)804 cm3T_2 = \frac{(603 \, \text{cm}^3) \times (400.15 \, \text{K})}{804 \, \text{cm}^3} T2=241290.45 K⋅cm3804 cm3T_2 = \frac{241290.45 \, \text{K} \cdot \text{cm}^3}{804 \, \text{cm}^3} T2≈300.11 KT_2 \approx 300.11 \, \text{K}
  5. Convert the final temperature back to Celsius.

    T2 (∘C)=T2 (K)−273.15T_2 \, (^\circ\text{C}) = T_2 \, (\text{K}) - 273.15 T2 (∘C)=300.11 K−273.15T_2 \, (^\circ\text{C}) = 300.11 \, \text{K} - 273.15 T2 (∘C)≈26.96 ∘CT_2 \, (^\circ\text{C}) \approx 26.96 \, ^\circ\text{C}

    The temperature at which the volume of the gas will be 603 cm3603 \, \text{cm}^3 is approximately 27.0 ∘C27.0 \, ^\circ\text{C}.

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