Question

1. what is the pressure of 1.20 mol of gas in a 10.0 l container at 310 k?

Explanation:

Step1: Identify the formula

We use the Ideal Gas Law, \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant (\( 0.0821 \, \text{L·atm/(mol·K)} \)), and \( T \) is temperature in Kelvin. Given \( n = 1.20 \, \text{mol} \), \( V = 10.0 \, \text{L} \) (assuming "10.0L" from the image), \( T = 310 \, \text{K} \).

Step2: Rearrange for \( P \)

Rearrange the formula to \( P = \frac{nRT}{V} \).

Step3: Substitute values

Substitute \( n = 1.20 \), \( R = 0.0821 \), \( T = 310 \), \( V = 10.0 \) into the formula:
\( P = \frac{1.20 \times 0.0821 \times 310}{10.0} \)

Step4: Calculate

First, calculate the numerator: \( 1.20 \times 0.0821 = 0.09852 \); \( 0.09852 \times 310 = 30.5412 \).
Then divide by \( V \): \( \frac{30.5412}{10.0} = 3.05412 \, \text{atm} \approx 3.05 \, \text{atm} \) (or check units if \( V \) was different, but assuming 10.0L).

Answer:

The pressure is approximately \( \boldsymbol{3.05 \, \text{atm}} \) (or adjust based on exact \( V \) and \( T \) precision).