10.0 grams of a gas occupies 12.5 liters at a pressure of 42.0 mm Hg. What is the volume when the pressure has increased to 75.0 mm Hg? 0.143 L 6.72 L 7.00 L 22.3 L

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 42.0 x 12.5 / 75.0

V2 = 7.0 L

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raquecgomez raquecgomez · Answer 2 · 2019-11-25T16:43:59+01:00

Answer:

7.00 L

Explanation:

The only thing that varies between the two situations is pressure and volume.

we have the ideal gas equation

[tex]PV=nrT[/tex]

We know that n = moles of substance remain constant, also the temperature and n corresponding to the ideal gas constant

Situation N1

[tex]P_1= 42.0 mmHg\\V_1= 12.5 L \\P_1V_1=nrT[/tex]

Situation N2

[tex]P_2= 75mmHg\\V_2= ?\\P_2V_2=nrT[/tex]

As nrT are equal both times, therefore we can match this term in both equations

[tex]Ecuation N1\\P_1V_1=nrT\\Ecuation N2\\ P_2V_2=nrT\\[/tex]

We equate both equations

[tex]P_2V_2=P_1V_1\\V_2=\frac{P_1V_1}{P_2}[/tex]

[tex]V_2=\frac{42mmHg.12.5 L}{75.0 mmHg} \\V_2= 7.00L[/tex]