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The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K10? Express the temperature to the nearest integer. View Available Hint(s) T1 T 1 T_1 = nothing ∘C Part B The molecules in an ideal gas at 10∘C have a root-mean-square (rms) speed vrms. At what temperature T2 (in degrees Celsius) will the molecules have twice the rms speed, 2vrms? Express the temperature to the nearest integer. View Available Hint(s) T2 T 2 T_2 = nothing ∘C

Respuesta :

Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

A)

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

T1 = 566 k = 293°C

B)

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

At

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

2√283 = √T2

Squaring on both sides:

(4)(283) = T2

T2 = 1132 k = 859°C

A) The temperature at which the average kinetic energy will have a value of 2K10 is; T1 = 293 °C

B) The temperature at which the molecules have twice the rms speed, 2vrms is; T2 = 859 °C

A) We are given;

Initial temperature; T = 10°C = 283 K

Initial kinetic energy; KE = K10

Final kinetic energy; KE1 = 2K10

Now,formula for average kinetic energy of the molecules of an ideal gas is given as;

KE = (3/2)kT

Where;

k is Boltzmann constant

T is temperature

We are told that in the second case, KE = 2K10. Thus;

2K10 = (3/2)kT1

K10 = ¾kT1 - - - (eq 2)

In the first instance, we have;

K10 = (3/2)kT - - - (eq 1)

Put (3/2)kT for K10 in eq 2 to get;

(3/2)kT = ¾kT1

k will cancel out to get;

(3/2)T = ¾T1

Make T1 the subject to get;

T1 = 2T

Thus;

T1 = 2 × 283

T1 = 566 K

Converting to °C gives;

T1 = 293 °C

B) We want to find the temperature T2 (in degrees Celsius) at which the molecules will have twice the rms speed, 2v_rms.

Formula for kinetic energy is also;

KE = ½mv²

Thus;

½m(v_rms)² = (3/2)kT

v_rms = √(3kT/m) - - - (eq 1)

When rms speed is 2v_rms, we have;

½m(2v_rms)² = (3/2)kT2

v_rms = √(¾kT2/m) - - - (eq 2)

Thus;

√(3kT/m) = √(¾kT2/m)

Square both sides to get;

(3kT/m) = (¾kT2/m)

4T = T2

T2 = 283 × 4

T2 = 1132 K

Converting to °C gives;

T2 = 859 °C

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