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25 June, 18:41

A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of volume of 90mL and pressure of 3.20 atm

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  1. 25 June, 19:09
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    The new temperature of the gas is 604K

    Explanation:

    Assuming standard temperature and pressure, to calculate the temperature of the gas we use the general gas equation;

    Step 1 : write the general gas equation

    P1V1 / T1 = P2V2/T2

    Step 2: Write out the values, covert the necessary values to the standard values.

    P1 = 0.800atm.

    V1 = 0.180L

    T1 = 29°C = 273 + 29 = 302K

    P2, 3.20atm

    V2 = 90mL = 90 * 10^-3L = 0.09L

    Step 3: Solve for T2

    The new temperature T2 of the gas is:

    T2 = P2V2T1 / P1V1

    T2 = 3.20 * 0.09 * 302 / 0.800 * 0.180

    T2 = 86.976 / 0.144

    T2 = 604K

    The new temperature is of the gas is 604K.
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