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mina [271]
3 years ago
10

What is the relationship between temperature and pressure pls​

Chemistry
2 answers:
love history [14]3 years ago
6 0

Answer:

When the temperature of a system goes up, the pressure also goes up, and vice versa. The relationship between pressure and temperature of a gas is stated by the Gay-Lussac's law.

Harman [31]3 years ago
3 0

Answer:

We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

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A mixture containing nitrogen and hydrogen weighs 3.48 g and occupies a volume of 7.47 L at 296 K and 1.02 atm. Calculate the ma
Sonbull [250]

Answer:

there is 2% of hydrogen and 98% of nitrogen (mass percent)

Explanation:

assuming ideal gas behaviour

P*V=n*R*T

n= P*V/(R*T)

where P= pressure=1.02 atm , V=volume=7.47 L , T=absolute temperature= 296 K and R= ideal gas constant = 0.082 atm*L/(mole*K)

thus

n= P*V/(R*T) = 1.02 atm*7.47 L/( 296 K * 0.082 atm*L/(mole*K)) = 0.314 moles

since the number of moles is related with the mass m through the molecular weight M

n=m/M

thus denoting 1 as hydrogen and 2 as nitrogen

m₁+m₂ = mt (total mass)

m₁/M₁+m₂/M₂ = n

dividing one equation by the other and denoting mass fraction w₁= m₁/mt , w₂= m₂/mt , w₂= 1- w₁

w₁/M₁+w₂/M₂ = n/mt

w₁/M₁+(1-w₁) /M₂ = n/mt

w₁*(1/M₁- 1/M₂) + 1/M₂ = n/mt

w₁=  (n/mt- 1/M₂) /(1/M₁- 1/M₂)

replacing values

w₁=  (n/mt- 1/M₂) /(1/M₁- 1/M₂) = (0.314 moles/3.48 g - 1/(14 g/mole)) /(1/(1 g/mole)-1/(14 g/mole))= 0.02 (%)

and w₂= 1-w₁= 0.98 (98%)

thus there is 2% of hydrogen and 98% of nitrogen

4 0
3 years ago
3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?
miv72 [106K]

Answer:

0.737 L

Explanation:

Charles law states for a fixed amount of gas, volume of the gas is directly proportional to the absolute temperature of the gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at the second instance

substituting the values

3.50 L / 727.0 K = V2 / 153.0 K

V2 = 0.737 L

new volume at 153.0 K is 0.737 L

8 0
4 years ago
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anastassius [24]

this is the answer

sorry if the answer is wrong

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3 years ago
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