Standard conditions: Temperature -0^circ C Pressure -101.325kPa Example 4. At 25^circ C and a pressure of 99.3 kPa (745 mmHg) a certain amount of a gas occupies a volume of .152 ml. Calculat the volume that the same amount of the gas will occupy at 0^circ C anc 1. a pressure of 101.33 kPa. 32.The pressure of bar (a) gas in a closed vessel is 96.0 kPa at 7^circ C . What will the pressure be if the vessel is cooled to 2. -33^circ C 33. In standard conditions, 1 g of air occupies a volume of 773 ml. What volume will the same mass of air occupy 3. at 0^circ C and a pressure of 93.3kPa(700mmHg) __ 37. A certain amount of a gas occupies a volume of 608 ml at a pressure of 98 .7 kPa and a temperature of 91^circ C . Find "57.What I __ volume is occupied by 27times 10^21 molecules of a 5.gas in standard conditions? __ 63. Determine e the volume occupied by 0.07 kg of N_(s) at 6. 24^circ C and 142 kPa (1065 mmHg).

Let's go through each problem step by step:<br /><br />### Example 4<br />Given:<br />- Initial temperature, \( T_1 = 25^{\circ}C = 298.15K \)<br />- Initial pressure, \( P_1 = 99.3 kPa \)<br />- Initial volume, \( V_1 = 0.152 \, L \)<br />- Final temperature, \( T_2 = 0^{\circ}C = 273.15K \)<br />- Final pressure, \( P_2 = 101.33 kPa \)<br /><br />Using the combined gas law:<br />\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]<br /><br />Solving for \( V_2 \):<br />\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]<br />\[ V_2 = \frac{99.3 \times 0.152 \times 273.15}{101.33 \times 298.15} \]<br />\[ V_2 \approx 0.132 \, L \]<br /><br />### Problem 32<br />Given:<br />- Initial pressure, \( P_1 = 96.0 kPa \)<br />- Initial temperature, \( T_1 = 7^{\circ}C = 280.15K \)<br />- Final temperature, \( T_2 = -33^{\circ}C = 240.15K \)<br /><br />Using Charles's Law:<br />\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]<br /><br />Solving for \( P_2 \):<br />\[ P_2 = \frac{P_1 T_2}{T_1} \]<br />\[ P_2 = \frac{96.0 \times 240.15}{280.15} \]<br />\[ P_2 \approx 81.6 kPa \]<br /><br />### Problem 33<br />Given:<br />- Mass of air, \( m = 1 \, g \)<br />- Volume at standard conditions, \( V_1 = 773 \, mL \)<br />- Standard pressure, \( P_1 = 101.325 kPa \)<br />- Final temperature, \( T_2 = 0^{\circ}C = 273.15K \)<br />- Final pressure, \( P_2 = 93.3 kPa \)<br /><br />Using the combined gas law:<br />\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]<br /><br />Solving for \( V_2 \):<br />\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]<br />\[ V_2 = \frac{101.325 \times 773 \times 273.15}{93.3 \times 273.15} \]<br />\[ V_2 \approx 773 \, mL \]<br /><br />### Problem 37<br />Given:<br />- Initial volume, \( V_1 = 608 \, mL \)<br />- Initial pressure, \( P_1 = 98.7 kPa \)<br />- Initial temperature, \( T_1 = 91^{\circ}C = 364.15K \)<br />- Final temperature, \( T_2 = 0^{\circ}C = 273.15K \)<br /><br />Using Charles's Law:<br />\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]<br /><br />Solving for \( V_2 \):<br />\[ V_2 = \frac{V_1 T_2}{T_1} \]<br />\[ V_2 = \frac{608 \times 273.15}{364.15} \]<br />\[ V_2 \approx 457 \, mL \]<br /><br />### Problem 57<br />Given:<br />- Number of molecules, \( N = 27 \times 10^{21} \)<br />- Standard pressure, \( P = 101.325 kPa \)<br />- Standard temperature, \( T = 273.15K \)<br />- Avogadro's number, \( N_A = 6.022 \times 10^{23} \)<br /><br />Using the ideal gas law:<br />\[ PV = nRT \]<br /><br />First, find the number of moles:<br />\[ n = \frac{N}{N_A} \]<br />\[ n = \frac{27 \times 10^{21}}{6.022 \times 10^{23}} \]<br />\[ n \approx 4.49 \times 10^{-2} \, moles \]<br /><br />Now, find the volume:<br />\[ V = \frac{nRT}{P} \]<br />\[ V = \frac{4.49 \times 10^{-2} \times 8.314 \times 273.15