A compressor takes 0.50m^3 of a gas at 33^circ C and 760 mm Hg and compresses it to 0.10m^3 cooling it to -55^circ C at the same time What is the pressure of the gas at these new conditions? 2,700 mm Hg 68 mm Hg 1.0 mm Hg 3.7times 10^-4 mm Hg

To find the pressure of the gas at the new conditions, we can use the combined gas law, which relates the initial and final states of a gas:<br /><br />\[<br />\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}<br />\]<br /><br />where:<br />- \( P_1 \) is the initial pressure (760 mm Hg),<br />- \( V_1 \) is the initial volume (0.50 m³),<br />- \( T_1 \) is the initial temperature in Kelvin,<br />- \( P_2 \) is the final pressure (unknown),<br />- \( V_2 \) is the final volume (0.10 m³),<br />- \( T_2 \) is the final temperature in Kelvin.<br /><br />First, convert the temperatures from Celsius to Kelvin:<br />- Initial temperature: \( T_1 = 33^\circ C + 273.15 = 306.15 \, K \)<br />- Final temperature: \( T_2 = -55^\circ C + 273.15 = 218.15 \, K \)<br /><br />Now, plug the known values into the combined gas law equation:<br /><br />\[<br />\frac{760 \, \text{mm Hg} \times 0.50 \, \text{m}^3}{306.15 \, \text{K}} = \frac{P_2 \times 0.10 \, \text{m}^3}{218.15 \, \text{K}}<br />\]<br /><br />Solve for \( P_2 \):<br /><br />\[<br />P_2 = \frac{760 \times 0.50 \times 218.15}{306.15 \times 0.10}<br />\]<br /><br />Calculate:<br /><br />\[<br />P_2 = \frac{760 \times 0.50 \times 218.15}{30.615}<br />\]<br /><br />\[<br />P_2 = \frac{82957}{30.615}<br />\]<br /><br />\[<br />P_2 \approx 2708.9 \, \text{mm Hg}<br />\]<br /><br />Rounding to the nearest whole number, the pressure of the gas at the new conditions is approximately 2709 mm Hg. The closest answer choice is:<br /><br />2,700 mm Hg