You have 0.5 Lof air at 203 K in an expandable container at constant pressure. You heat the container to 273 K. What is the volume of air? (V_(1))/(T_(1))=(V_(2))/(T_(2)) T_(1)=203K V_(2)=? T_(2)=273K

To solve this problem, we can use the ideal gas law equation:<br /><br />$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$<br /><br />Given:<br />$T_{1} = 203K$<br />$V_{1} = 0.5L$<br />$T_{2} = 273K$<br /><br />We need to find $V_{2}$.<br /><br />Substituting the given values into the equation:<br /><br />$\frac{0.5L}{203K} = \frac{V_{2}}{273K}$<br /><br />Cross-multiplying:<br /><br />$0.5L \times 273K = V_{2} \times 203K$<br /><br />Solving for $V_{2}$:<br /><br />$V_{2} = \frac{0.5L \times 273K}{203K} \approx 0.674L$<br /><br />Therefore, the volume of air in the container when heated to 273K is approximately 0.674L.