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malfutka [58]
3 years ago
6

If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder will change from 7

0 kilopascals to ...
Chemistry
1 answer:
Elza [17]3 years ago
6 0

Answer:

P₂ = 140 KPa

Explanation:

Given data:

Initial volume = 8.0 L

Final volume = 4.0 L

Initial pressure = 70 KPa

Final pressure = ?

Solution:

According to Boyle's law

P₁V₁ = P₂V₂

P₂ = P₁V₁  / V₂

P₂ = 70 KPa ×8.0 L/4.0 L

P₂ = 560 KPa .L / 4.0 L

P₂ = 140 KPa

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Above question is incomplete. Complete question is attached below
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7 0
3 years ago
Is a compound a metal or non metal
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Answer:

Compounds between Nonmetals and Nonmetals

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3 0
3 years ago
A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t
soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0
2 years ago
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