How ,any moles of an ideal gas is present in a 0.45 L sample of gas st 273 K and .50 atm pressure?

0.00027 mol
99.6 mol
0.01 mol
50.37 mol

1 answer

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

\[ PV = nRT \]

Where:

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

Given:

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

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

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

Now substituting the values into the equation:

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

Calculating the denominator:

\[ R \cdot T = 0.0821 \cdot 273 \approx 22.4143 , \text{L} \cdot \text{atm} / \text{mol} \]

Now substitute back into the equation for \( n \):

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

Calculating the numerator:

\[ 0.50 \cdot 0.45 = 0.225 \]

Now:

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

Rounding gives \( n \approx 0.01 , \text{mol} \).

Therefore, the number of moles of the ideal gas present in the sample is approximately 0.01 mol.