Question

Q.9. A compressed-air tank holds 0.5 m^3 air at a temperature of 285 K and a pressure of 880 kPa. What volume would the air occupy if it were released into the atmosphere, where the pressure is 101 kPa and the temperature is 303 K?
The answer is 4.63 m ^3( I only need the method)​

Answer 1

Another statement is, "Volume is directly proportional to the number of moles."

The volume increases as the number of moles increases. It does not depend on the sizes or the masses of the molecules.

V

∝

n

, where  

V

is the volume, and  

n

is the number of moles.

V

n

=

k

, where  

k

is a proportionality constant.

We can rewrite this as

V

1

n

1

=

V

2

n

2

Equal volumes of hydrogen, oxygen, or carbon dioxide contain the same number of molecules.

STP is 0 °C and 1 bar.

One mole of an ideal gas occupies 22.71 L at STP. Thus, its molar volume at STP is 22.71 L

Example Problem

A 6.00 L sample at 25.0 °C and 2.00 atm contains 0.500 mol of gas. If we add 0.250 mol of gas at the same pressure and temperature, what is the final total volume of the gas?

Solution

The formula for Avogadro's law is:

V

1

n

1

=

V

2

n

2

V

1

=

6.00 L

;

n

1

=

0.500 mol

V

2

=

?

;

m

m

l

n

2

=

0.500 mol + 0.250 mol = 0.750 mol

V

2

=

V

1

×

n

2

n

1

V

2

=

6.00 L

×

0.750

mol

0.500

mol

=

9.00 L

Ernest Z. ·  135 · Feb 27 2014

Question #779c9

Gay-Lussac’s Law is an ideal gas law where at constant volume, the pressure of an ideal gas is directly proportional to its absolute temperature. In other words, Gay-Lussac's Law states that the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins.

Simplified, this means that if you increase the temperature of a gas, the pressure rises proportionally. Pressure and temperature will both increase or decrease simultaneously as long as the volume is held constant.

The law has a simple mathematical form if the temperature is measured on an absolute scale, such as in kelvins. The Gay-Lussac’s Law is expressed as:

P

1

T

1

=  

P

2

T

2

Where  

P

1

stands for the initial pressure of the gas,  

T

1

stands for the initial temperature,  

P

2

stands for the final pressure of the gas, and  

T

2

stands for the final temperature.

This law holds true because temperature is a measure of the average kinetic energy of a substance; when the kinetic energy of a gas increases, its particles collide with the container walls more rapidly and exert more pressure.

Take a sample of gas at STP 1 atm and 273 K and double the temperature.

1

a

t

m

273

K

=

P

546

K

546

a

t

m

K

273

K

=

P

P = 2 atm

Doubling the temperature, likewise doubled the pressure.

Answer 2

Answer:

4.63m^3

Explanation:

$P_{1} =880000Pa\\P_{2} =101000Pa\\V_{1} =0.5m^{3} \\V_{2}= ?\\T_{1}=285K\\T_{2}=303K$

We first start with the formula $\frac{P_{1}V_{1} }{T_{1} } =\frac{P_{2}V_{2}}{T_{2}}$.

After doing a litle algebra, we will get the formula $\frac{T_{2} \frac{P_{1}V_{1} }{T_{1} } }{P_{2} } =V_{2}$.

Now we submit the variables and get the answer.