18a) Crtate Charler law. vtater that the volume of a given mar of a gak is directly proportional to its abrolute temperature at a constant prossure. b The volume of a sample of ndrogen gas at a tomperatu of agik and 1.0 times 10^5 Parale was 3.5 times 10^-2 mathrm(~m)^3 . Calculato the temperature at which the volume of legan would be 2.5 times 10^-2 mathrm(~m)^3 at 1.0 times 105 Pascaln

To solve this problem, we can use Charles's Law, which states that the volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure. Mathematically, this can be expressed as:<br /><br />\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]<br /><br />where:<br />- \( V_1 \) is the initial volume<br />- \( T_1 \) is the initial temperature<br />- \( V_2 \) is the final volume<br />- \( T_2 \) is the final temperature<br /><br />Given:<br />- \( V_1 = 3.5 \times 10^{-2} \, \mathrm{m}^3 \)<br />- \( T_1 = 300 \, \mathrm{K} \) (assuming the initial temperature is 300 K)<br />- \( V_2 = 2.5 \times 10^{-2} \, \mathrm{m}^3 \)<br />- \( P = 1.0 \times 10^5 \, \mathrm{Pa} \) (constant pressure)<br /><br />We need to find \( T_2 \).<br /><br />Using Charles's Law:<br /><br />\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]<br /><br />Rearranging to solve for \( T_2 \):<br /><br />\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \]<br /><br />Substituting the given values:<br /><br />\[ T_2 = \frac{2.5 \times 10^{-2} \, \mathrm{m}^3 \cdot 300 \, \mathrm{K}}{3.5 \times 10^{-2} \, \mathrm{m}^3} \]<br /><br />\[ T_2 = \frac{7.5 \times 10^{-2} \, \mathrm{m}^3 \cdot \mathrm{K}}{3.5 \times 10^{-2} \, \mathrm{m}^3} \]<br /><br />\[ T_2 = \frac{7.5}{3.5} \, \mathrm{K} \]<br /><br />\[ T_2 \approx 2.14 \, \mathrm{K} \]<br /><br />Therefore, the temperature at which the volume of the gas would be \( 2.5 \times 10^{-2} \, \mathrm{m}^3 \) at \( 1.0 \times 10^5 \, \mathrm{Pa} \) is approximately \( 2.14 \, \mathrm{K} \).