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neonofarm [45]
3 years ago
12

A volume of gas at 1.10 atm was measured at 507 ml. What will be the volume if the pressure is adjusted to 1.90

Chemistry
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

294mL

Explanation:

We can use this equation to find the unknown volume:

P_1V_1=P_2V_2

To solve, plug in the values we have and solve for the unknown value:

(1.10atm)(507mL)=(1.90atm)V_2\\\frac{(1.10atm)(507mL)}{1.90atm} =V_2\\V_2=293.53mL

Rounding to three significant figures gives you the value of 294mL

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Answer:

d=0.92\frac{kg}{m^{3}}

Explanation:

Using the Ideal Gas Law we have PV=nRT and the number of moles n could be expressed as n=\frac{m}{M}, where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

PV=\frac{m}{M}RT

If we pass the V to divide:

P=\frac{m}{V}\frac{RT}{M}

As the density is expressed as d=\frac{m}{V}, we have:

P=d\frac{RT}{M}

Solving for the density:

d=\frac{PM}{RT}

Then we need to convert the units to the S.I.:

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P=1bar*\frac{0.98atm}{1bar}

P=0.98atm

M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}

M=0.0289\frac{kg}{mol}

Finally we replace the values:

d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}

d=9.2*10^{-4}\frac{kg}{L}

d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}

d=0.92\frac{kg}{m^{3}}

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Answer:

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