A300cmcub sample of gas exert pressure of 600kpa at 27degree c what pressure would it exert in 200cmcub container 20 degree c

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=27 ∘C+273.15=300.15 KT_1 = 27 \, ^\circ\text{C} + 273.15 = 300.15 \, \text{K} T2=20 ∘C+273.15=293.15 KT_2 = 20 \, ^\circ\text{C} + 273.15 = 293.15 \, \text{K}
  2. Identify the given values.

    Initial Volume (V1V_1) = 300 cm3300 \, \text{cm}^3
    Initial Pressure (P1P_1) = 600 kPa600 \, \text{kPa}
    Initial Temperature (T1T_1) = 300.15 K300.15 \, \text{K}
    Final Volume (V2V_2) = 200 cm3200 \, \text{cm}^3
    Final Temperature (T2T_2) = 293.15 K293.15 \, \text{K}
    Final Pressure (P2P_2) = ?

  3. Apply the Combined Gas Law.

    The Combined Gas Law states that for a fixed amount of gas, the ratio of the product of pressure and volume to the absolute temperature is constant:

    P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

    To find P2P_2, rearrange the formula:

    P2=P1V1T2T1V2P_2 = \frac{P_1 V_1 T_2}{T_1 V_2}
  4. Substitute the values and calculate P2P_2.

    P2=(600 kPa)×(300 cm3)×(293.15 K)(300.15 K)×(200 cm3)P_2 = \frac{(600 \, \text{kPa}) \times (300 \, \text{cm}^3) \times (293.15 \, \text{K})}{(300.15 \, \text{K}) \times (200 \, \text{cm}^3)} P2=52767000 kPa⋅cm3⋅K60030 K⋅cm3P_2 = \frac{52767000 \, \text{kPa} \cdot \text{cm}^3 \cdot \text{K}}{60030 \, \text{K} \cdot \text{cm}^3} P2≈878.99 kPaP_2 \approx 878.99 \, \text{kPa}

    The pressure the gas would exert is approximately 879 kPa.

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