Question

0.65 moles of O2 originally at 85°C is cooled

Answer 1

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:

$pV=nRT$

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

$T=40C+273=313 K$ is the temperature of the gas

$R=0.082 atm L mol^{-1} K^{-1}$ is the gas constant

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

$p=\frac{nRT}{V}=\frac{(0.65)(0.082)(313)}{8.0}=2.09 atm$