Arigid cylinder contains 2.00 liters of gas at a temperature of 25°c. if the pressure of this gas is changed from 1.00 atmospheres to 0.250 atmospheres, what will be the new temperature (in kelvin, reported to three significant figures) of the gas? (the volume is constant.)

- The new temperature would be 74.5 K.

Solution:- Volume is constant and according to Gay Lussac's law, "at constant volume, pressure is directly proportional to the kelvin temperature."

=25+273=298K

= ?

= 1.00 atm

= 0.250 atm

Let's plug in the values in the equation:

= 74.5K

The new temperature , T2 = 74.5 K

There are several gas equations in various processes:

PV = NkT

N = number of gas particles

n = number of moles

R = gas constant (8,31.10 ^ 3 J / kmole K

k = Boltzmann constant (1,38.10⁻²³)

n =  N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / M

m = mass

M = relative molecular mass

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules

So it applies: the ratio of gas volume will be equal to the ratio of gas moles

At a constant temperature, the gas volume is inversely proportional to the pressure applied

When the gas pressure is kept constant, the gas volume is proportional to the temperature

When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature

Combined with Boyle's law and Gay Lussac's law

P1 = initial gas pressure (N / m² or Pa)

V1 = initial gas volume (m³)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

In the problem, the conditions that are set constant are volume, so we use Gay Lussac's Law

A rigid cylinder contains 2.00 liters of gas at a temperature of 25 ° c. the pressure of this gas is changed from 1.00 atm to 0.250 atm and the volume is constant

So:

V = constant = 2 L

T1 = 25 ° C  = 298 K

P1 = 1 atm

P2 = 0.25 atm

Then:

a description of Charles’s law

link

Charles's law

link

State Boyle's, Charles's, and Gay-Lussac's laws

link

astronomers such as edwin hubble.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).