A sample of helium gas collected at a pressure of 1.11 atm and a temperature of 10.0°C is found to occupy a volume of 23.9 liters. How many moles of He gas are in the sample?

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### Example Problem - Calculating Moles of Gas

#### Problem Statement:
"A sample of helium gas collected at a pressure of 1.11 atm and a temperature of 10.0°C is found to occupy a volume of 23.9 liters. How many moles of He gas are in the sample?"

#### Solution:

To solve for the number of moles of helium gas, we can use the Ideal Gas Law equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atmospheres, atm)
- \( V \) = volume (in liters, L)
- \( n \) = number of moles (mol)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin, K)

First, convert the temperature from Celsius to Kelvin:
\[ T(K) = T(°C) + 273.15 \]
\[ T(K) = 10.0 + 273.15 \]
\[ T(K) = 283.15 \text{ K} \]

Given:
- \( P = 1.11 \text{ atm} \)
- \( V = 23.9 \text{ L} \)
- \( R = 0.0821 \text{ L·atm/(K·mol)} \)
- \( T = 283.15 \text{ K} \)

Substitute these values into the Ideal Gas Law equation and solve for \( n \):

\[ (1.11 \text{ atm}) (23.9 \text{ L}) = n (0.0821 \text{ L·atm/(K·mol)}) (283.15 \text{ K}) \]

\[ 26.529 \text{ atm·L} = n (23.238115 \text{ L·atm/mol}) \]

\[ n = \frac{26.529}{23.238115} \]

\[ n \approx 1.142 \text{ mol} \]

Thus, there are approximately 1.142 moles of helium gas in the sample.
Transcribed Image Text:### Example Problem - Calculating Moles of Gas #### Problem Statement: "A sample of helium gas collected at a pressure of 1.11 atm and a temperature of 10.0°C is found to occupy a volume of 23.9 liters. How many moles of He gas are in the sample?" #### Solution: To solve for the number of moles of helium gas, we can use the Ideal Gas Law equation: \[ PV = nRT \] Where: - \( P \) = pressure (in atmospheres, atm) - \( V \) = volume (in liters, L) - \( n \) = number of moles (mol) - \( R \) = ideal gas constant (0.0821 L·atm/(K·mol)) - \( T \) = temperature (in Kelvin, K) First, convert the temperature from Celsius to Kelvin: \[ T(K) = T(°C) + 273.15 \] \[ T(K) = 10.0 + 273.15 \] \[ T(K) = 283.15 \text{ K} \] Given: - \( P = 1.11 \text{ atm} \) - \( V = 23.9 \text{ L} \) - \( R = 0.0821 \text{ L·atm/(K·mol)} \) - \( T = 283.15 \text{ K} \) Substitute these values into the Ideal Gas Law equation and solve for \( n \): \[ (1.11 \text{ atm}) (23.9 \text{ L}) = n (0.0821 \text{ L·atm/(K·mol)}) (283.15 \text{ K}) \] \[ 26.529 \text{ atm·L} = n (23.238115 \text{ L·atm/mol}) \] \[ n = \frac{26.529}{23.238115} \] \[ n \approx 1.142 \text{ mol} \] Thus, there are approximately 1.142 moles of helium gas in the sample.
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