PLEASE HELP

The pressure of a gas varies jointly with the amount of the gas (measured in moles) and the temperature and inversely with the volume of the gas.
If the pressure is 1,218 kPa (kilo Pascals) when the number of moles is 8, the temperature is 290* Kelvin, and the volume is 960 cc,
find the pressure when the number of moles is 6, the temperatire is 250° K, and the volume is 360 cc.

this means, when you put more gas into the same volume at the same temperature, the pressure goes up linearly. and it goes down linearly, when you remove gas.

when you increase the temperature of the same amount of gas in the same volume, the pressure goes up linearly. and it goes down linearly when you decrease the temperature.

when you decrease the volume for the same amount of gas at the same temperature, the pressure goes up linearly. and it goes down linearly, if you increase the volume.

now, all 3 attributes are changed.

and so, the pressure changes as a combination of all 3 factors.

the moles decrease from 8 to 6. that is a factor of 6/8 = 3/4.

the temperature decreases from 290°K to 250°K. that is a factor of 250/290 = 25/29.

the volume decreases from 960 cc to 360 cc. this is an (inverse) factor of 960/360 = 16/6 = 8/3.

so, the pressure is

1,218 × 3/4 × 25/29 × 8/3 = 1,218 × 25/29 × 2 =

= 2,100 kPa

semsee45 semsee45 · Answer 2 · 2022-08-22T10:13:04+02:00

Answer:

2100 kPa

Step-by-step explanation:

The pressure of a gas varies jointly with the amount of the gas and the temperature and inversely with the volume:

[tex]\implies P \propto \dfrac{nT}{V}[/tex]

where:

P = pressure (measured in kilo Pascals, kPa).
n = number of moles.
T = temperature (measured in kelvins, K).
V = volume (measured in cubic centimeters, cc).

[tex]\textsf{If }a \propto b, \textsf{ then } a=kb \textsf{ for some constant } k:[/tex]

[tex]\implies P =\dfrac{knT}{V}[/tex]

Given:

P = 1218 kPa
n = 8 mol
T = 290 K
V = 960 cc

Substitute the given values into the derived equation to find the constant of variation (k):

[tex]\implies 1218 =\dfrac{k(8)(290)}{960}[/tex]

[tex]\implies 1218 =\dfrac{2320k}{960}[/tex]

[tex]\implies 1169280=2320k[/tex]

[tex]\implies k=\dfrac{1169280}{2320}[/tex]

[tex]\implies k=504[/tex]

Substitute the found value of k into the equation:

[tex]\implies P =\dfrac{504nT}{V}[/tex]

To find the pressure (P) when:

n = 6 mol
T = 250 K
v = 360 cc

substitute the given values into the equation and solve for P:

[tex]\implies P =\dfrac{504(6)(250)}{360}[/tex]

[tex]\implies P=\dfrac{756000}{360}[/tex]

[tex]\implies P=2100\:\: \sf kPa[/tex]