A gas has an initial volume of 8.0 L, e pressure of 2.0 atm , and a temperature of soo K. If the gas expands to 5,0 L and the temperature increases to 400 K , what is the new pressure of the gas?

To solve this problem, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation:<br /><br />\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]<br /><br />where \( P_1 \) is the initial pressure, \( V_1 \) is the initial volume, \( T_1 \) is the initial temperature, \( P_2 \) is the final pressure, \( V_2 \) is the final volume, and \( T_2 \) is the final temperature.<br /><br />Given:<br />- Initial volume, \( V_1 = 8.0 \, \text{L} \)<br />- Initial pressure, \( P_1 = 2.0 \, \text{atm} \)<br />- Initial temperature, \( T_1 = 800 \, \text{K} \)<br />- Final volume, \( V_2 = 50 \, \text{L} \)<br />- Final temperature, \( T_2 = 400 \, \text{K} \)<br /><br />We need to find the final pressure, \( P_2 \).<br /><br />Rearranging the ideal gas law to solve for \( P_2 \):<br /><br />\[ P_2 = P_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \]<br /><br />Substituting the given values:<br /><br />\[ P_2 = 2.0 \, \text{atm} \times \frac{8.0 \, \text{L}}{50 \, \text{L}} \times \frac{400 \, \text{K}}{800 \, \text{K}} \]<br /><br />\[ P_2 = 2.0 \, \text{atm} \times \frac{8.0}{50} \times \frac{1}{2} \]<br /><br />\[ P_2 = 2.0 \, \text{atm} \times 0.16 \times 0.5 \]<br /><br />\[ P_2 = 0.16 \, \text{atm} \]<br /><br />Therefore, the new pressure of the gas is \( 0.16 \, \text{atm} \).