Question

How many moles of an ideal gas is present in a 0.45 L sample of gas at 273 K and 0.50 atm pressure?(1 point)Responses50.37 mol50.37 mol0.01 mol0.01 mol0.00027 mol0.00027 mol99.6 mol99.6 mol

Answer 1

To calculate the number of moles of an ideal gas, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

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

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

We can solve for \( n \) (number of moles):

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

Substituting in the values:

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

Calculating the denominator:

\[ 0.0821 \times 273 \approx 22.4143 \, \text{L·atm/(K·mol)} \]

Now calculating \( n \):

\[ n = \frac{0.50 \times 0.45}{22.4143} \]

\[ n = \frac{0.225}{22.4143} \approx 0.01 \, \text{mol} \]

Therefore, the number of moles of the ideal gas present in a 0.45 L sample at 273 K and 0.50 atm pressure is approximately **0.01 mol**.

The correct response is **0.01 mol**.