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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