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One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com pressed from 2 atm to 3 atm by a reversible isothermal path. Subsequantly, it is expanded back to 2 atm by a reversible adiabatic path. The value of final state in litre is
- a)13.67
- b)31.1
- c)18.1
- d)20.1
Correct answer is option 'A'. Can you explain this answer?
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One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com press...
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One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com press...
Given:
- One mole of an ideal gas
- Cv = 1.5 R (where R is the molar gas constant)
- Initial pressure (Pi) = 2 atm
- Final pressure (Pf) = 3 atm
- Temperature (T) = 400 K
To find:
The final state of the gas in liters.
Solution:
Step 1: Calculating the initial volume (Vi)
Using the ideal gas law equation, we can calculate the initial volume (Vi) of the gas at Pi and T.
The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = molar gas constant
T = temperature
Rearranging the equation, we get:
V = (nRT) / P
Substituting the given values:
Vi = (1 mol * 2 atm * 0.0821 L/mol K * 400 K) / 2 atm
Vi = 32.84 L
Therefore, the initial volume (Vi) of the gas is 32.84 L.
Step 2: Calculating the final volume (Vf) using the isothermal compression
Since the compression is isothermal, the temperature remains constant throughout the process. Therefore, we can use the following equation to calculate the final volume (Vf) of the gas:
(Vi / Vf) = (Pf / Pi)
Substituting the given values:
(32.84 L / Vf) = (3 atm / 2 atm)
Cross-multiplying and solving for Vf:
2 * 32.84 L = 3 * Vf
Vf = (2 * 32.84 L) / 3
Vf = 21.89 L
Therefore, the final volume (Vf) of the gas after isothermal compression is 21.89 L.
Step 3: Calculating the final volume (Vf) using the adiabatic expansion
Since the expansion is adiabatic, there is no heat exchange with the surroundings. Therefore, we can use the following equation to calculate the final volume (Vf) of the gas:
(Pi * Vi^γ) = (Pf * Vf^γ)
Where:
γ = Cp / Cv (ratio of specific heat capacities)
Given that Cv = 1.5 R, we can calculate γ as follows:
γ = Cp / Cv
γ = (Cv + R) / Cv
γ = (1.5 R + R) / 1.5 R
γ = 2.5
Substituting the given values:
(2 atm * 32.84 L^2.5) = (3 atm * Vf^2.5)
Simplifying the equation:
65.68 L^2.5 = 3 atm * Vf^2.5
Taking the power of 2/5 on both sides:
(65.68 L^2.5)^(2/5) = (3 atm * Vf^2.5)^(2/5)
Simplifying further:
65.68 L^(5/5) = 3 atm^(2/5) * Vf^(
- One mole of an ideal gas
- Cv = 1.5 R (where R is the molar gas constant)
- Initial pressure (Pi) = 2 atm
- Final pressure (Pf) = 3 atm
- Temperature (T) = 400 K
To find:
The final state of the gas in liters.
Solution:
Step 1: Calculating the initial volume (Vi)
Using the ideal gas law equation, we can calculate the initial volume (Vi) of the gas at Pi and T.
The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = molar gas constant
T = temperature
Rearranging the equation, we get:
V = (nRT) / P
Substituting the given values:
Vi = (1 mol * 2 atm * 0.0821 L/mol K * 400 K) / 2 atm
Vi = 32.84 L
Therefore, the initial volume (Vi) of the gas is 32.84 L.
Step 2: Calculating the final volume (Vf) using the isothermal compression
Since the compression is isothermal, the temperature remains constant throughout the process. Therefore, we can use the following equation to calculate the final volume (Vf) of the gas:
(Vi / Vf) = (Pf / Pi)
Substituting the given values:
(32.84 L / Vf) = (3 atm / 2 atm)
Cross-multiplying and solving for Vf:
2 * 32.84 L = 3 * Vf
Vf = (2 * 32.84 L) / 3
Vf = 21.89 L
Therefore, the final volume (Vf) of the gas after isothermal compression is 21.89 L.
Step 3: Calculating the final volume (Vf) using the adiabatic expansion
Since the expansion is adiabatic, there is no heat exchange with the surroundings. Therefore, we can use the following equation to calculate the final volume (Vf) of the gas:
(Pi * Vi^γ) = (Pf * Vf^γ)
Where:
γ = Cp / Cv (ratio of specific heat capacities)
Given that Cv = 1.5 R, we can calculate γ as follows:
γ = Cp / Cv
γ = (Cv + R) / Cv
γ = (1.5 R + R) / 1.5 R
γ = 2.5
Substituting the given values:
(2 atm * 32.84 L^2.5) = (3 atm * Vf^2.5)
Simplifying the equation:
65.68 L^2.5 = 3 atm * Vf^2.5
Taking the power of 2/5 on both sides:
(65.68 L^2.5)^(2/5) = (3 atm * Vf^2.5)^(2/5)
Simplifying further:
65.68 L^(5/5) = 3 atm^(2/5) * Vf^(
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One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com pressed from 2 atm to 3 atm by a reversible isothermal path. Subsequantly, it is expanded back to 2 atm by a reversible adiabatic path. The value of final state in litre isa)13.67b)31.1c)18.1d)20.1Correct answer is option 'A'. Can you explain this answer?
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One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com pressed from 2 atm to 3 atm by a reversible isothermal path. Subsequantly, it is expanded back to 2 atm by a reversible adiabatic path. The value of final state in litre isa)13.67b)31.1c)18.1d)20.1Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com pressed from 2 atm to 3 atm by a reversible isothermal path. Subsequantly, it is expanded back to 2 atm by a reversible adiabatic path. The value of final state in litre isa)13.67b)31.1c)18.1d)20.1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal ga s Cv = 1.5 R at temperature 40 0K is com pressed from 2 atm to 3 atm by a reversible isothermal path. Subsequantly, it is expanded back to 2 atm by a reversible adiabatic path. The value of final state in litre isa)13.67b)31.1c)18.1d)20.1Correct answer is option 'A'. Can you explain this answer?.
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