Asample of gas initially has a volume of 859 ml at 565 k and 2.20 atm. what pressure will the sample have if the volume changes to 268 ml while the temperature is increased to 815 k?

We have that the pressure the sample have if the volume changes to 268 ml while the temperature is increased to 815 k is

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A sample of gas initially has a volume of 859 ml at 565 k and 2.20 atm. What pressure will the sample have if the volume changes to 268 ml while the temperature is increased to 815 k

Generally the equation for the ideal gas   is mathematically given as

PV=nRT

Where

Therefore

THEREFORE

the pressure the sample have if the volume changes to 268 ml while the temperature is increased to 815 k is

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10.2 atm

Explanation:

Use ideal gas law:

PV = nRT

Initial number of moles is:

(2.20 atm) (0.859 L) = n (0.0821 atm L / mol / K) (565 K)

n = 0.0407 mol

At the new volume and temperature, the pressure is:

P (0.268 L) = (0.0407 mol) (0.0821 atm L / mol / K) (815 K)

P = 10.2 atm

a. carbon monoxide from car exhaust

b.particulate matter, such as soot

c.water vapor from a factory smokestack

d.acid rain

these are the answer choices im on that question right now