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blah asked in Science & MathematicsWeather · 8 years ago

A balloon is floating around outside your window. The temperature outside is 27 , and the air pressure is 0.80?

1.) A balloon is floating around outside your window. The temperature outside is 27 degrees celcius , and the air pressure is 0.800atm . Your neighbor, who released the balloon, tells you that he filled it with 3.00moles of gas. What is the volume of gas inside this balloon?

2)A 20.0L gas cylinder is filled with 8.60moles of gas. The tank is stored at 13 degrees celcius . What is the pressure in the tank?

3) A 260L. kiln is used for vitrifying ceramics. It is currently operating at 1375 degrees celcius, and the pressure is 0.9250atm. How many moles of air molecules are within the confines of the kiln?

4) A 10.0L gas cylinder has been filled with 5.00moles of gas. You measure the pressure to be 3.00atm . What is the temperature inside the tank?

5) A 31.0g sample of nitrogen, N2, has a volume of 60.0L and a pressure of 610mmHg. . What is the temperature, in kelvins and degrees Celsius, of the gas?

6) A gas mixture contains oxygen and argon at partial pressures of 0.50atm and 445 mmHg. If nitrogen gas added to the sample increases the total pressure to 1200torr , what is the partial pressure in torr of the nitrogen added?

2 Answers

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  • ?
    Lv 6
    8 years ago
    Favorite Answer

    ideal gas law which is PV=nRT

    n=moles of gas

    V=volume (L)

    P=pressure (kPa)

    T=temperature (in Kelvin)

    R=ideal gas constant which is 8.314 J K-1 mol-1

    1.) A balloon is floating around outside your window. The temperature outside is 27 degrees celcius , and the air pressure is 0.800atm . Your neighbor, who released the balloon, tells you that he filled it with 3.00moles of gas. What is the volume of gas inside this balloon?

    so question 1. solve for V

    rearrange equation also (27 degrees=300.15K) also standard atmosphere to kilo pascals so

    (0.8 atm=81.06 kPa)

    V= (nRT)/P

    V=(3x8.314x300.15)/81.06

    V= 92.356L

    2)A 20.0L gas cylinder is filled with 8.60moles of gas. The tank is stored at 13 degrees celcius . What is the pressure in the tank?

    Question 2

    solve for P

    rearrange equation also (13 degrees = 286.15K)

    P = (nRT)/V

    P = (8.6 x 8.314 x 286.15) / 20

    P = 1022.99 kPa

    3) A 260L. kiln is used for vitrifying ceramics. It is currently operating at 1375 degrees celcius, and the pressure is 0.9250atm. How many moles of air molecules are within the confines of the kiln?

    Question 3

    Solve for n also (1375 degrees =1648.15K) (0.925atm = 93.725625kPa)

    Soo n = (PV)/(RT)

    n = (93.725625 x 260)/ (8.314 x 1648.15)

    n = 1.778 moles

    4) A 10.0L gas cylinder has been filled with 5.00moles of gas. You measure the pressure to be 3.00atm . What is the temperature inside the tank?

    Question 4

    Solve for T (3 atm = 303.975kPa)

    T= (PV) / (nR)

    T = (303.975 x 10) / (5 x 8.314)

    T = 73.124 K

    5) A 31.0g sample of nitrogen, N2, has a volume of 60.0L and a pressure of 610mmHg. . What is the temperature, in kelvins and degrees Celsius, of the gas?

    Solve for T and moles (610mmHg = 81.32666kPa)

    Molar mass of nitrogen = 14.00672

    Number of moles = (# of grams) ÷ (molar mass) (31g of nitrogen = 2.2132 moles)

    T= (PV) / (nR)

    T = (81.32666 x 60) / (2.2132 x 8.314)

    T = 265.19 K or -7.9622 degrees

    6) A gas mixture contains oxygen and argon at partial pressures of 0.50atm and 445 mmHg. If nitrogen gas added to the sample increases the total pressure to 1200torr , what is the partial pressure in torr of the nitrogen added?

    Oxygen (0.5 atm = 380 torr (mmHg))

    Argon (445 mmHg)

    1200 – (445 +380) = 375 mmHg

    Nitrogen = 375 mmHg

    I better get all this right!!

  • Anonymous
    5 years ago

    Truthfully speakme, i am not as sufferer as you, to appear in any respect these things... I simply can inform the reply. My answer is sure. Its a bent of a scorching air to upward push.. This does not mean that it'll now not rise if the encompassing air is of the identical temperature. When its surrounding temperature is of the equal temperature, there could be no transfer of heat, for this reason the temperature will remain regular. This may hold the scorching air arised. If the temperature round it cools down, there would be a transfer of warmness, so as to calm down the air, hence sinking it.

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