Question

1. a gas has 0.75 mol at 2.00 atm and 273k

what is its volume?

Explanation:

Step1: Recall Ideal Gas Law

The Ideal Gas Law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant, \( T \) is temperature. \( R = 0.0821 \, \text{L·atm/(mol·K)} \) or \( 0.0821 \, \text{dm}^3·\text{atm/(mol·K)} \) (since \( 1 \, \text{L} = 1 \, \text{dm}^3 \)).

Step2: Identify Given Values

\( P = 2.00 \, \text{atm} \), \( n = 0.75 \, \text{mol} \), \( T = 273 \, \text{K} \), \( R = 0.0821 \, \text{dm}^3·\text{atm/(mol·K)} \).

Step3: Rearrange Formula for \( V \)

From \( PV = nRT \), solve for \( V \): \( V = \frac{nRT}{P} \).

Step4: Substitute Values

\( V = \frac{0.75 \, \text{mol} \times 0.0821 \, \text{dm}^3·\text{atm/(mol·K)} \times 273 \, \text{K}}{2.00 \, \text{atm}} \).

First, calculate numerator: \( 0.75 \times 0.0821 \times 273 \approx 0.75 \times 22.4133 \approx 16.81 \).

Then divide by \( P \): \( V \approx \frac{16.81}{2.00} \approx 8.405 \, \text{dm}^3 \) (or L).

Answer:

The volume of the gas is approximately \( 8.41 \, \text{dm}^3 \) (or \( 8.41 \, \text{L} \)).