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For 2 mol of CO2 gas at 300 K
- a)Translational kinetic energy is 1800 cal
- b)Ratio of rotational to vibrational kinetic energy is (all degrees of freedom are activated) 1 : 2
- c)Ratio of Urms to Umps is 3 : 2
- d)Slope of (log P) vs (log V) curve is log (600 R).
Correct answer is option 'A'. Can you explain this answer?
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For 2 mol of CO2gas at 300 Ka)Translational kinetic energy is 1800 cal...
Transtational K. E. =3/2 nRT
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For 2 mol of CO2gas at 300 Ka)Translational kinetic energy is 1800 cal...
Translational Kinetic Energy:
The translational kinetic energy of a gas molecule is given by the formula:
KE = (3/2)RT
where KE is the kinetic energy, R is the gas constant, and T is the temperature in Kelvin.
Given that there are 2 moles of CO2 gas at 300 K, we can calculate the translational kinetic energy as follows:
KE = (3/2)RT
= (3/2)(8.314 J/mol·K)(300 K)
≈ 18,810 J/mol
To convert this to calories, we can use the conversion factor:
1 cal = 4.184 J
So, the translational kinetic energy is approximately:
18,810 J/mol × (1 cal/4.184 J)
≈ 4494 cal/mol
Therefore, the correct answer for part (a) is 4494 cal.
Ratio of Rotational to Vibrational Kinetic Energy:
For a diatomic molecule like CO2, there are three degrees of rotational freedom and two degrees of vibrational freedom. When all degrees of freedom are activated, the ratio of rotational to vibrational kinetic energy is 1:2. This is because rotational motion contributes less to the total kinetic energy compared to vibrational motion.
Therefore, the correct answer for part (b) is 1:2.
Ratio of Urm to Umps:
The ratio of the root mean square velocity (Urm) to the most probable speed (Umps) can be calculated using the formula:
Urm/Umps = √(8RT/πM)
where M is the molar mass of the gas molecule.
For CO2, the molar mass is approximately 44 g/mol. Plugging in the values, we get:
Urm/Umps = √(8RT/πM)
= √(8(8.314 J/mol·K)(300 K)/(π(44 g/mol)))
≈ √(1990.343/137.43)
≈ √(14.486)
≈ 3.805
Therefore, the correct answer for part (c) is 3:2.
Slope of log(P) vs log(V) Curve:
The slope of the log(P) vs log(V) curve can be determined using the ideal gas law equation:
PV = nRT
Taking the logarithm of both sides:
log(PV) = log(nRT)
Using logarithmic properties, we can rewrite this equation as:
log(P) + log(V) = log(nRT)
This equation represents a straight line with a slope of 1. Therefore, the slope of the log(P) vs log(V) curve is 1.
Therefore, the correct answer for part (d) is log(1) = 0.
The translational kinetic energy of a gas molecule is given by the formula:
KE = (3/2)RT
where KE is the kinetic energy, R is the gas constant, and T is the temperature in Kelvin.
Given that there are 2 moles of CO2 gas at 300 K, we can calculate the translational kinetic energy as follows:
KE = (3/2)RT
= (3/2)(8.314 J/mol·K)(300 K)
≈ 18,810 J/mol
To convert this to calories, we can use the conversion factor:
1 cal = 4.184 J
So, the translational kinetic energy is approximately:
18,810 J/mol × (1 cal/4.184 J)
≈ 4494 cal/mol
Therefore, the correct answer for part (a) is 4494 cal.
Ratio of Rotational to Vibrational Kinetic Energy:
For a diatomic molecule like CO2, there are three degrees of rotational freedom and two degrees of vibrational freedom. When all degrees of freedom are activated, the ratio of rotational to vibrational kinetic energy is 1:2. This is because rotational motion contributes less to the total kinetic energy compared to vibrational motion.
Therefore, the correct answer for part (b) is 1:2.
Ratio of Urm to Umps:
The ratio of the root mean square velocity (Urm) to the most probable speed (Umps) can be calculated using the formula:
Urm/Umps = √(8RT/πM)
where M is the molar mass of the gas molecule.
For CO2, the molar mass is approximately 44 g/mol. Plugging in the values, we get:
Urm/Umps = √(8RT/πM)
= √(8(8.314 J/mol·K)(300 K)/(π(44 g/mol)))
≈ √(1990.343/137.43)
≈ √(14.486)
≈ 3.805
Therefore, the correct answer for part (c) is 3:2.
Slope of log(P) vs log(V) Curve:
The slope of the log(P) vs log(V) curve can be determined using the ideal gas law equation:
PV = nRT
Taking the logarithm of both sides:
log(PV) = log(nRT)
Using logarithmic properties, we can rewrite this equation as:
log(P) + log(V) = log(nRT)
This equation represents a straight line with a slope of 1. Therefore, the slope of the log(P) vs log(V) curve is 1.
Therefore, the correct answer for part (d) is log(1) = 0.
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For 2 mol of CO2gas at 300 Ka)Translational kinetic energy is 1800 calb)Ratio of rotational to vibrational kinetic energy is (all degrees of freedom are activated) 1 : 2c)Ratio of Urmsto Umpsis 3 : 2d)Slope of (log P) vs (log V) curve islog (600 R).Correct answer is option 'A'. Can you explain this answer?
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