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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?

Respuesta :

MathPhys

Answer:

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

okpalawalter8

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

[tex]P_2=10atm[/tex]

From the question we are told

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

[tex]\frac{P1V1}{T1}=\frac{P2V2}{T2}[/tex]

Therefore

[tex]P_2=\frac{P1V1T2}{V2T1}\\\\P_2=\frac{2.20*859*815}{268*565}[/tex]

[tex]P_2=10atm[/tex]

THEREFORE

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

[tex]P_2=10atm[/tex]

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