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A 6.5 L sample of nitrogen at 25 degree C and 0.76 atm is allowed to expand to 1

ID: 2253031 • Letter: A

Question

A 6.5 L sample of nitrogen at 25 degree C and 0.76 atm is allowed to expand to 13.0 L. The tem­perature remains constant. What is the final pressure? 0.75 atm 0.12 atm 0.38 atm 0.063 atm 3.0 atm A balloon filled with helium gas has a volume of 699 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height? Answer in units of mL A gas has a pressure of 2.22 atm and occupies a volume of 7.8 L. If the gas is compressed to a volume of 2.03 L, what will its pressure be, assuming constant temperature? Answer in units of atm The piston of an internal combustion engine compresses 451 mL of gas. The final pressure is 15 times greater than the initial pressure. What is the final volume of the gas, assuming constant temperature? Answer in units of mL. A weather balloon with a volume of 3.04706 L is released from Earth's surface at sea level. What volume will the balloon occupy at an altitude of 20.0 km, where the air pressure is 10 kPa? Answer in units of L A balloon filled with oxygen gas occupies a volume of 4.7 L at 22 degree C. What volume will the gas occupy at 199 degree C? Answer in units of L A sample of nitrogen gas is contained in a piston with a freely moving cylinder. At 0 degree C, the volume of the gas is 213 mL. To what temperature must the gas be heated to occupy a volume of 593 mL? Answer in units of degree C If the Kelvin temperature of an ideal gas is doubled while maintaining a constant pres­sure, the pressure doubles. the volume increases by a factor of 4. the volume doubles. the volume is halved. A sample of a gas occupies 1418 mL at 321 degree C and 740 torr. At what temperature would it occupy 1194.6 mL if the pressure is kept constant? Answer in units of degree C

Explanation / Answer

1)

PV/T =constant

6.5 * 0.76 = 13 * P

P = 0.38 atm


2)

699 * 1 = 0.5 * V

V = 1398 ml


3)

7.8 * 2.22 = 2.03 * P

P = 8.53 atm


4)

P * 451 = 15 * P * V

V = 451 / 15 = 30.067 ml


6)

Pressure at surface = 101325

V = 3.0476 * 101325 / 100000 = 3.08 L


7)

4.7/(273+22) = V / (273 + 199)

V = 7.52 L


8)

213/273 = 593/(T+273)

T = 487 celcius


9)

volume doubles


10)

1418/(273+321) = 1194.6/(273+T)

T = 227.418 celcius

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