Period __ Gas Laws Daily Assignment Chates' Law Boyle's Law Arogadro's Law Dalton's Law (V_(1))/(T_(1))=(V_(2))/(T_(2)) P_(1)V_(1)=P_(2)V_(2) (V_(1))/(n_(1))=(V_(2))/(n_(2)) P_(T)=P_(1)+P_(2)+P_(3)+P_(3) 1. A sample of nitrogen gas is 35.0^circ C occupies 5.86 L. Find the new volume if the temperature is increased to 72.0^circ C Assume pressure is constant. 2. Ifa balloon has a volume of 7.08 ml and a pressure of 414,34 kPa, what will its pressure be if its volume decreases to 1.57 ml. (Temperature remains constant) 3. A ball has a gas volume of 7.64 L and pressure of 1.65 atm. How large will the volume be if we increase the pressure to 3.16 atm.Temperature remains constant.

1. To solve this problem, we can use Charles' Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure remains constant. The formula for Charles' Law is:<br /><br />$\frac{V_1}{T_1} = \frac{V_2}{T_2}$<br /><br />where $V_1$ is the initial volume, $T_1$ is the initial temperature, $V_2$ is the final volume, and $T_2$ is the final temperature.<br /><br />Given:<br />$V_1 = 5.86$ L<br />$T_1 = 35.0^{\circ}C$<br />$T_2 = 72.0^{\circ}C$<br /><br />First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15:<br /><br />$T_1 = 35.0^{\circ}C + 273.15 = 308.15$ K<br />$T_2 = 72.0^{\circ}C + 273.15 = 345.15$ K<br /><br />Now, we can substitute the values into the formula and solve for $V_2$:<br /><br />$\frac{5.86}{308.15} = \frac{V_2}{345.15}$<br /><br />$V_2 = \frac{5.86 \times 345.15}{308.15} \approx 6.34$ L<br /><br />Therefore, the new volume of the nitrogen gas sample when the temperature is increased to $72.0^{\circ}C$ is approximately 6.34 L.<br /><br />2. To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume, provided the temperature remains constant. The formula for Boyle's Law is:<br /><br />$P_1V_1 = P_2V_2$<br /><br />where $P_1$ is the initial pressure, $V_1$ is the initial volume, $P_2$ is the final pressure, and $V_2$ is the final volume.<br /><br />Given:<br />$P_1 = 414.34$ kPa<br />$V_1 = 7.08$ ml<br />$V_2 = 1.57$ ml<br /><br />We can substitute the values into the formula and solve for $P_2$:<br /><br />$414.34 \times 7.08 = P_2 \times 1.57$<br /><br />$P_2 = \frac{414.34 \times 7.08}{1.57} \approx 1836.57$ kPa<br /><br />Therefore, the pressure of the balloon when its volume decreases to 1.57 ml is approximately 1836.57 kPa.<br /><br />3. To solve this problem, we can again use Boyle's Law. Given:<br />$P_1 = 1.65$ atm<br />$V_1 = 7.64$ L<br />$P_2 = 3.16$ atm<br /><br />We can substitute the values into the formula and solve for $V_2$:<br /><br />$1.65 \times 7.64 = 3.16 \times V_2$<br /><br />$V_2 = \frac{1.65 \times 7.64}{3.16} \approx 4.08$ L<br /><br />Therefore, the volume of the ball when the pressure is increased to 3.16 atm is approximately 4.08 L.