What pressure is required to compress an ideal gas from 10.0 ft^3 and 1.00 atm t

Question

What pressure is required to compress an ideal gas from 10.0 ft^3 and 1.00 atm to 2.00 ft^3 under isothermal conditions? A rubber balloon was filled with helium at 25 degree C and placed in a beaker of liquid nitrogen at -196 degree C. The volume of the cold helium was 15 cm^3. Assuming ideal gas behavior and isobaric conditions, what was the volume of the helium at 25degree C? According to Gay-Lussac's law of combining volumes, 2 L of hydrogen gas will react with 1 L of oxygen gas (measured at the same temperature and pressure) to produce steam. What volume of steam will be formed? A toy balloon originally held 1.00 g of helium gas and had a radius of 10.0 cm. During the night. 0.25 g of the gas effused from the balloon. Assuming ideal gas behavior under these constant pressure and temperature conditions, what was the radius of the balloon the next morning? What is the mass of air in a 250-mi. Erlenmever flask (actual volume is 267 mL) at a typical laboratory pressure of 715 torr and a temperature of 21 degree C? Assume air to be an ideal gas having an average molar mass of 29.0 g/mol. 6. How many molecules of gas are in a 0.535-L vacuum system at a pressure of 3.5 Times 10^-7 torr and 25 degree C? Assume ideal gas behavior. A 1.00-mol sample of O_2 and a 3.00-mol sample of H_2 are mixed isothermally in a 125.3-L container at 125 degree C. Assuming ideal gas behavior, calculate the partial pressure of each gas and the total pressure of the gaseous mixture. The pressure of a gaseous mixture of NH_3 and N_2 decreased from 1.50 kPa to 1.00 kPa after the NH_3 was absorbed from the mixture. Assuming ideal gas behavior, find the composition of the original mixture. A mixture of ethane and ethene occupied 35.5 L at 1.000 bar and 405 K. This mixture reacted completely with 110.3 g of O_2 to produce CO_2 and H_2O. What was the composition of the original mixture? Assume ideal gas behavior.

Explanation / Answer

1 )  

form bolyl's law P1 V1 = P2 V2

    1 x 10 = P2 x 2

P2 = 5 atm

pressure = 5 atm

2)

low temp :   T1 = 273 - 196 = 77 K

here volume = 15 cm^3

T2 = 25 + 273 = 298 K

V2 = how much ?

V1 / T1 = V2 / T2

15 / 77 = V2 / 298

V2 = 58.05 cm^3

volume = 58.05 cm^3

3)

2H2 + O2 --------------> 2 H2O

2         1                      2

so 2 litres steam formd

6 )

V = 0.535 L

P = 3.5 x 10^-7 torr

1 atm = 760 torr

??? = 3.5 x 10^-7 torr

P = 4.60 x 10^-10 atm

T = 273 + 25 = 298 K

P V = n R T

4.60 x 10^-19 x 0.535 = n x 0.0821 x 298

n = 1 x 10^-11

moles = 1.00 x 10^-11

1 mol -----------------> 6.023 x 10^23 molecules

1.0x 10^-11 moles ------------> x molecules

x = 6.06 x 10^12

number of molecules = 6.06 x 10^12

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