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A gas occupies 1 L at 27°C. If the pressure is held constant, what is its volume at 40°C? Use 273.15 for the conversion from Celsius to Kelvin.

Physics

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Answer

1.04 L\text{1.04 L}

Step-by-step solution

4 steps

Hey Yano, this problem involves Charles's Law, not Boyle's Law, because it relates volume and temperature. Boyle's Law relates pressure and volume. The notation "v²" likely refers to the final volume, V2V_2.

  1. State 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. The formula is:

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

    Rearranging to solve for V2V_2:

    V2=V1×T2T1V_2 = V_1 \times \frac{T_2}{T_1}
  2. Convert temperatures from Celsius to Kelvin.

    Gas law calculations require temperatures to be in Kelvin.

    TK=TC+273.15T_K = T_C + 273.15

    Given:
    V1=1 LV_1 = 1 \text{ L}
    T1=27∘CT_1 = 27^\circ \text{C}
    T2=40∘CT_2 = 40^\circ \text{C}

    Convert T1T_1:

    T1=27∘C+273.15=300.15 KT_1 = 27^\circ \text{C} + 273.15 = 300.15 \text{ K}

    Convert T2T_2:

    T2=40∘C+273.15=313.15 KT_2 = 40^\circ \text{C} + 273.15 = 313.15 \text{ K}
  3. Substitute the values into the Charles's Law formula.

    V2=1 L×313.15 K300.15 KV_2 = 1 \text{ L} \times \frac{313.15 \text{ K}}{300.15 \text{ K}}
  4. Calculate the final volume, V2V_2.

    V2=1 L×1.043305...V_2 = 1 \text{ L} \times 1.043305... V2≈1.04 LV_2 \approx 1.04 \text{ L}

    The final volume is:

    1.04 L\text{1.04 L}
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