Question

chapter 5 - gases examination
started: nov 23 at 9:31pm
quiz instructions
the fifth exam. this is an examination on the material in the gases chapter of the text.
question 7
4 pts
convert 5.0atm of ideal gas in a 5.0l container at 260k into moles.

Explanation:

Step1: Recall the ideal gas law

The ideal gas law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. We need to solve for \( n \), so we rearrange the formula to \( n=\frac{PV}{RT} \).

Step2: Identify the values

We are given:

• \( P = 5.0 \, \text{atm} \)
• \( V = 5.0 \, \text{L} \)
• \( T = 260 \, \text{K} \)
• The ideal gas constant \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \) (this is the value for \( R \) when pressure is in atm, volume in L, temperature in K, and moles in mol).

Step3: Substitute the values into the formula

Substitute \( P = 5.0 \, \text{atm} \), \( V = 5.0 \, \text{L} \), \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \), and \( T = 260 \, \text{K} \) into \( n=\frac{PV}{RT} \):

$$ n=\frac{(5.0 \, \text{atm})(5.0 \, \text{L})}{(0.0821 \, \frac{\text{L·atm}}{\text{mol·K}})(260 \, \text{K})} $$

Step4: Calculate the numerator and the denominator

First, calculate the numerator: \( (5.0 \, \text{atm})(5.0 \, \text{L}) = 25.0 \, \text{L·atm} \)

Then, calculate the denominator: \( (0.0821 \, \frac{\text{L·atm}}{\text{mol·K}})(260 \, \text{K}) = 0.0821\times260 \, \frac{\text{L·atm}}{\text{mol}} \approx 21.346 \, \frac{\text{L·atm}}{\text{mol}} \)

Step5: Divide the numerator by the denominator

Now, divide the numerator by the denominator:

$$ n=\frac{25.0 \, \text{L·atm}}{21.346 \, \frac{\text{L·atm}}{\text{mol}}} \approx 1.17 \, \text{mol} $$

Answer:

\( \approx 1.2 \, \text{mol} \) (or more precisely \( \approx 1.17 \, \text{mol} \))