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NARA [144]
3 years ago
15

what would happen to the volume of 10L carbon dioxide, if the pressure is doubled and temperature is constant?​

Chemistry
1 answer:
lara [203]3 years ago
5 0

Answer:

The volume will goes to decrease.

Explanation:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

when pressure is going to increase the volume will goes to decrease and vice versa.

Consider the initial volume is 10 L at initial pressure of 15 atm when pressure is increased to 30 atm the volume will be,

Now we will put the values in formula,

P₁V₁ = P₂V₂

15 atm × 10 L = 30  atm × V₂

V₂ = 150 atm. L/ 30 atm

V₂ = 5 L

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B) A company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you
Yuri [45]

Answer:

43.1atm is the pressure using gas law and 27.2atm using Van der Waals Law.

Explanation:

Ideal gas law is:

PV = nRT

<em>Where P is pressure in atm</em>

<em>V is volume = 4.00L</em>

<em>n are moles of the gas (For chlorine Molar Mass: 70.90g/mol):</em>

<em>500g * (1mol / 70.90g) = 7.052 moles</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 25°C + 273 = 298K</em>

To solve the pressure, P:

P = nRT/V

P = 7.052mol*0.082atmL/molK*298K / 4.00L

P = 43.1atm is the pressure using gas law.

Van der Waals equation is:

P + a(\frac{n}{V})^2 * (V-nb) = nRT

<em>Where a is 6.58L²atm*mol⁻²</em>

<em>b = 0.056Lmol⁻²</em>

Solving for pressure:

P + a(\frac{n}{V})^2 = \frac{nRT}{(V-nb)}

P  = \frac{nRT}{(V-nb)}-a(\frac{n}{V})^2

P  = \frac{7.052mol*0.082atmL/molK*298K}{(4.00L-7.052mol*0.056L*mol)}-6.58L^2mol^{-2}(\frac{7.052mol}{4.00L})^2

P = 172.323 / 3.6051 - 20.4866

P = 27.2atm using Van der Waals Law

8 0
3 years ago
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