Question

How many moles of an ideal gas are present in a 0.45 L sample of gas at 273 K and 0.50 atm pressure?
A. 50.37 mol B. 0.00027 mol C. 0.01 mol D. 99.6 mol

Answer 1

To determine the number of moles of an ideal gas present in a 0.45 L sample at 273 K and 0.50 atm pressure, we can use the Ideal Gas Law, which is given by:

\[ PV = nRT \]

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

Given:
- \( P = 0.50 \, \text{atm} \)
- \( V = 0.45 \, \text{L} \)
- \( T = 273 \, \text{K} \)

Now, we rearrange the Ideal Gas Law to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the known values into the equation:

\[ n = \frac{(0.50 \, \text{atm}) (0.45 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})) (273 \, \text{K})} \]

Calculating the denominator:

\[ R \cdot T = 0.0821 \times 273 \approx 22.4143 \]

Now calculate \( n \):

\[ n = \frac{(0.50) (0.45)}{22.4143} \]

Calculating the numerator:

\[ 0.50 \times 0.45 = 0.225 \]

Now, dividing by the denominator:

\[ n \approx \frac{0.225}{22.4143} \approx 0.01003 \, \text{mol} \]

Thus, the number of moles of the gas is approximately 0.01 mol.

The correct answer is **C. 0.01 mol**.