If a gas with a pressure of 6 atmospheres is cooled from 500 K to 250 K then what is its final pressure if the volume does not change? 20833 atm 30 atm 3 atm 0.33 atm

## Step 1<br />The problem involves the application of Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature, provided the volume remains constant. This can be mathematically represented as:<br />### \( \frac{P1}{T1} = \frac{P2}{T2} \)<br />where \( P1 \) and \( T1 \) are the initial pressure and temperature, and \( P2 \) and \( T2 \) are the final pressure and temperature.<br /><br />## Step 2<br />In this problem, the initial pressure \( P1 \) is 6 atmospheres, the initial temperature \( T1 \) is 500 K, the final temperature \( T2 \) is 250 K, and we are asked to find the final pressure \( P2 \).<br /><br />## Step 3<br />Substitute the given values into the Gay-Lussac's Law formula:<br />### \( \frac{6}{500} = \frac{P2}{250} \)<br /><br />## Step 4<br />Solve for \( P2 \) by cross-multiplying and dividing:<br />### \( P2 = \frac{6 \times 250}{500} = 3 \) atmospheres