What is the pressure exerted by 3.00 mol of ar gas with a volume of 50.0 l at 25.0ºc (298.2 k)?

A rigid cylinder with a movable piston contains a sample of gas. At 300. K, this sample has a pressure of 240. Kilopascals and a

iren [92.7K]

Answer:

The volume of this sample when the temperature is changed to 150 K and the pressure is changed to 160 kPa is 52.5 mL.

Explanation:

Boyle's law says that: "The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure" and is expressed mathematically as:

P * V = k

where k is a constant.

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure, by means of a constant of proportionality that is applied directly. So Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$\frac{V}{T}=k$

Gay-Lussac's law states that the pressure of a fixed volume of a gas is directly proportional to its temperature. In other words, if the volume of a certain quantity of ideal gas remains constant, the quotient between pressure and temperature remains constant:

$\frac{P}{T}=k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T}=k$

Considering an initial state 1 and a final state 2, it is satisfied:

$\frac{P1*V1}{T1}=\frac{P2*V2}{T2}$

In this case:

P1: 240 kPa
V1: 70 mL
T1: 300 K
P2: 160 kPa
V2: ?
T2: 150 K

Replacing:

$\frac{240 kPa*70 mL}{300 K}=\frac{160 kPa*V2}{150 K}$

Solving:

$V2=\frac{150 K}{160 kPa} *\frac{240 kPa*70 mL}{300 K}$

V2= 52.5 mL

The volume of this sample when the temperature is changed to 150 K and the pressure is changed to 160 kPa is 52.5 mL.

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koban [17]

Answer:

Hi,

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Explanation:

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Answer:

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Explanation:

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