ellie420f
contestada

0.65 moles of O2 originally at 85°C is cooled
such that it now occupies 8.0 L at 40.°C.
What is the final pressure exerted by the gas?

Respuesta :

skyluke89

Answer:

2.09 atm

Explanation:

We can solve this problem by using the equation of state for an ideal gas, which relates the pressure, the volume and the temperature of an ideal gas:

[tex]pV=nRT[/tex]

where

p is the pressure of the gas

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

In this problem we have:

n = 0.65 mol is the number of moles of the gas

V = 8.0 L is the final volume of the gas

[tex]T=40C+273=313 K[/tex] is the temperature of the gas

[tex]R=0.082 atm L mol^{-1} K^{-1}[/tex] is the gas constant

Solving for p, we find the final pressure of the gas:

[tex]p=\frac{nRT}{V}=\frac{(0.65)(0.082)(313)}{8.0}=2.09 atm[/tex]

dsdrajlin

0.65 moles of O₂ at 40. °C and 2.1 atm occupy a volume of 8.0 L.

0.65 moles (n) of O₂ originally at 85°C is cooled. It occupies 8.0 L (V) at 40.°C (T). We will convert the final temperature to Kelvin using the following expression.

[tex]K = \°C + 273.15 = 40\°C + 273.15 = 313 K[/tex]

We can calculate the final pressure (P) exerted by the gas using the ideal gas equation.

[tex]P \times V = n \times R \times T\\\\P = \frac{n \times R \times T}{V} = \frac{0.65 mol \times (\frac{0.082atm.L}{mol.K} ) \times 313K}{8.0L} = 2.1 atm[/tex]

0.65 moles of O₂ at 40. °C and 2.1 atm occupy a volume of 8.0 L.

Learn more: https://brainly.com/question/4147359

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