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5 September, 09:34

A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What is the new pressure?

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  1. 5 September, 10:01
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    New pressure is 0.534 atm

    Explanation:

    Given:

    Initial volume of the gas, V₁ = 250 mL

    Initial pressure of the gas, P₁ = 1.00 atm

    Initial temperature of the gas, T₁ = 20° C = 293 K

    Final volume of the gas, V₂ = 500 mL

    Final pressure of the gas = P₂

    Final temperature of the gas, T₁ = 40° C = 313 K

    now,

    we know for a gas

    PV = nRT

    where,

    n is the moles

    R is the ideal gas constant

    also, for a constant gas

    we have

    (P₁V₁/T₁) = (P₂V₂/T₂)

    on substituting the values in the above equation, we get

    (1.00 * 250) / 293 = (P₂ * 500) / 313

    or

    P₂ = 0.534 atm

    Hence, the new pressure is 0.534 atm
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