Question

Draw a diagram for each of processes (isothermal, isobaric, isochoric) in variables (P, V), (P, T) and (V, T). Express density of an ideal gas using the equation of state: PV = n/M RT. Explain every step. One mode of oxygen gas is at a pressure of 6.00 and a temperature of 27.0 degree C. If the gas is heated at constant volume until the pressure triples, what is the final temperature? If the gas is heated so that both the volume and pressure are doubled, what is the final temperature? HW an ideal gas occupies a volume of 1.0 cm^3 at 20 degree C and atmospheric pressure. Determine the number of molecules of gas in the container. If the pressure is reduced to 1.0 times 10^22 Pa while the temperature constant, how many moles of gas remain in the container? HW Gas is confined in a link at a pressure of 11.0 atm and a temperature of 25.0 degree C. If two thirds of the gas is withdrawn and the temperature is raised to 75.0 degree C, what is the new pressure in the tank? Gas is obtained in an 8.00 L vessel at a temperature of 20.0 degree C and a pressure of 9.00 atm. Determine the number of moles of gas in the vessel. How many molecules are in the vessel? A weather balloon is designed to expand to a maximum radium of 20 m at the altitude where the pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at the atmospheric pressure and 300 K_1, what is its radium at liftoff? The density of helium gas as 0 degree C is = 0.179kg/m^3. The temperature is then raised to T = 100 degree C, but the pressure is kept constant. Considering helium as an ideal gas, calculate the new density of the gas. Read textbook, starting on pager 402, Get ready for the quiz. solve problems ##31, 33 above To be ready for the quiz, you need to know the macroscopic parameters of a gas Understand the physical properties of an ideal gas, understand the specifies of isothermal, isobaric and isochoric and isochoric processes, Understand the equation of state.

31 and 33

Answer 1

Answer 31)

(a) Given: P1= 1 atm = 1.013x10⁵ Pa, T1 = 20°C =T, V1 =1.0 cm³ = V₂ =V.

Find: (a) N1 = n1/N_A , N_A = 6.02 molecules/mole. (b) Given: P2 = 1.0 x 10^-11 Pa , T2 = T1 = 20°C =T = T,

V2 =V1 = V= 1.0 cm³ .

Find: n2

Conversions V2 =V1 = V= 1.0 cm³ 1.0 [cm]³ = 1.0 [cm( )m/cm]³ = 1.0 [cm(10^-2)m/cm]³ = 1.0 [(10^-2)m]³ = 1.0 x10^-6 m .

T2 = T1 = T = 20°C = 273.15 K + 20 = 293.15 K.

Physical Principle Ideal Gas Law: PV = nRT ------------eq(1)

R = the Universal Gas Constant = 8.31 J/mole K.

(a)

T1 = T = 293.15 K T and V1 = V = 1.0 x10^-6 m

n1 = P1V1/(RT1) = P1V1/(RT) ---------------eq(2)

= 1.013x10^-5 Pa *1.0 x10^-6 m³ /(8.31 J/mole K*293.15 K) = 4.2x10^-5 moles

N1 = n1/NA = 4.2x10^-5 moles/6.02 molecules/mole = 2.5x10¹⁹ molecules.

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(b) Use Eq. (2) now for n2 with T2 = T = 293.15 K T and V2 = V = 1.0 x10-6 m gives

n2 = P₂V₂/(RT₂ ) = P₂V/(RT)-------------------------eq (3)

= 1.0 x 10^-11 Pa x 1.0 x10^-6 m³ /(8.31 J/mole K*293.15 K)

=4.1 x10^-21 mol.

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Answer 33)

let there be (n) gram mole of gas in tank initially
P₁V = n R T₁

75° C=348K

25° C=298K
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2n/3 moles is released, so left in tank in (n/3) moles, V = tank same
P₂V = (n/3) R T₂
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divide
[P₂/P₁] = [1/3][T₂/T₁] = (1/3)[348 K/298K] = 0.3893
P2 = 0.3893*P₁ = 11*0.3893 = 4.28 atm