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17 December, 03:38

A 7.70 L 7.70 L container holds a mixture of two gases at 47 ° C. 47 °C. The partial pressures of gas A and gas B, respectively, are 0.344 atm 0.344 atm and 0.893 atm. 0.893 atm. If 0.190 mol 0.190 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

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  1. 17 December, 03:59
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    1.884 atm

    Explanation:

    We have at first, a mix of two gases A and B, with a pressure and temperature. The total pressure of this mix is the sum of both pressures fo these gases so:

    P = 0.344 + 0.893 = 1.237 atm

    Now after this, a third gas is added but there is no change in temperature or volume, which means that this gas may have a behavior of ideal gas, therefore, we can use the expression for ideal gas which is:

    P = nRT/V

    R is the gas constant, in this case is 0.082 L atm / K mol

    Replacing all the values that we have of volume and temperature, we can calculate the pressure that this gas exert:

    P = 0.19 * 0.082 * (47+273.15) / 7.7

    P = 0.647 atm

    Now the total pressure, we add this value and we have:

    P' = 1.237 + 0.647

    P' = 1.884 atm
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