Chemistry, 14.07.2019 08:30 quan91

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.)

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Answer from: mhzperfect

- 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."

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

T_1 =25+273=298K

T_2 = ?

P_1 = 1.00 atm

P_2 = 0.250 atm

Let's plug in the values in the equation:

$\frac{1.00}{0.250}=\frac{298}{T_2}$

$T_2=\frac{298*0.250}{1.00}$

T_2 = 74.5K

Answer from: raquelqueengucci25

The new temperature , T2 = 74.5 K

Further explanation

There are several gas equations in various processes:

1. The ideal ideal gas equation PV = nRT

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

2. Avogadro's hypothesis

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

V1: V2 = n1: n2 3. Boyle's Law

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

p1.V1 = p2.V2 4. Charles's Law

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

V1 / T1 = V2 / T2 5. Gay Lussac's Law

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

P1 / T1 = P2 / T2 6. Law of Boyle-Gay-Lussac

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

P1.V1 / T1 = P2.V2 / T2

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:

$\rm \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\\dfrac{1\:atm}{298\:K\:C}= \dfrac{0.25\:atm}{T_2}\\\\T_2=\dfrac{0.25\:atm\times 298}{1\:atm}\\\\T_2=74.5\:K$

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Arigid cylinder contains 2.00 liters of gas at a temperature of 25°c. if the pressure of this gas is...

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