1) A balloon containing 1.75 miles if He gas has a volume of 0.500L at a given t

Question

1) A balloon containing 1.75 miles if He gas has a volume of 0.500L at a given temperature and pressure. If balloon loses some amount of helium gas without changing temperature and pressure and now contains 1.15 moles of He, what js the new volume lf the balloon in liters? Name the Gas law?
2) A can contains a sample of gas under a pressure of 1.85 atm at 28 degrees celsuis. The can is placed in a boiler room and pressure is measured as 2.64 atm. Calculate the temperature in (kelvin and Celsius) of the hot can. Name the Gas law? 1) A balloon containing 1.75 miles if He gas has a volume of 0.500L at a given temperature and pressure. If balloon loses some amount of helium gas without changing temperature and pressure and now contains 1.15 moles of He, what js the new volume lf the balloon in liters? Name the Gas law?
2) A can contains a sample of gas under a pressure of 1.85 atm at 28 degrees celsuis. The can is placed in a boiler room and pressure is measured as 2.64 atm. Calculate the temperature in (kelvin and Celsius) of the hot can. Name the Gas law? Name the Gas law?
2) A can contains a sample of gas under a pressure of 1.85 atm at 28 degrees celsuis. The can is placed in a boiler room and pressure is measured as 2.64 atm. Calculate the temperature in (kelvin and Celsius) of the hot can. Name the Gas law?

Explanation / Answer

(1) Avogadro's Law states that the volume occupied by an ideal gas is directly proportional to the number of molecules of the gas present in the container.

V is proportional to n

So V / V' = n / n'

Where

V = initial volume = 0.500 L

V' = final volume = ?

n = initial number of moles = 1.75 moles

n' = final number of moles = 1.15 moles

Plug the values we get

V' = (Vn') / n

   = ( 0.500x1.15) / 1.75

   = 0.328 L

Therefore he new volume lf the balloon is 0.328 L

(2) Gay-Lussac's Law states that, for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

So    P / P' = T / T'

Where

P = initial pressure = 1.85 atm

P' = final pressure = 2.64 atm

T = initial temperature = 28oC = 28+273 = 301 K

T' = final temperature = ?

Plug the values we get

T' = (P'T) / P

   = ( 2.64x301) / 1.85

   = 429.5 K

= 429.5-273 oC

= 156.5 oC

Therefore the final temperature is 429.5 K     OR      156.5 oC

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