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A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845?
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A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This...
Solution:
Given data:
Initial volume of the gas, V1 = 10 m3
Initial pressure of the gas, P1 = 2 bar
Final pressure of the gas, P2 = 15 bar
The process is isothermal, which means the temperature of the gas remains constant during the process.
To find the work done during the process, we can use the formula for reversible isothermal work:
W = nRT ln(V2/V1)
Where n is the number of moles of gas, R is the gas constant, T is the temperature of the gas, and V2/V1 is the ratio of the final volume to the initial volume.
Since the process is isothermal, the temperature remains constant, so we can use the ideal gas law to find the number of moles of gas:
PV = nRT
n = PV/RT
Where P and V are the initial pressure and volume of the gas, respectively, and R is the gas constant.
Substituting the values, we get:
n = (2 bar) x (10 m3) / (8.314 J/mol-K) x (273 K)
n = 0.9568 moles
Now we can calculate the final volume of the gas using the ideal gas law:
PV = nRT
V2 = nRT/P2
Substituting the values, we get:
V2 = (0.9568 moles) x (8.314 J/mol-K) x (273 K) / (15 bar)
V2 = 0.142 m3
So the ratio of final volume to initial volume is:
V2/V1 = 0.142 m3 / 10 m3
V2/V1 = 0.0142
Now we can calculate the work done during the process using the reversible isothermal work formula:
W = nRT ln(V2/V1)
Substituting the values, we get:
W = (0.9568 moles) x (8.314 J/mol-K) x (273 K) x ln(0.0142)
W = -6,652 J
So the work required for the process is -6,652 J, which means work is done on the gas during the process.
Explanation:
In this question, we are given the initial volume and pressure of an ideal gas and asked to find the work required to compress the gas in a reversible isothermal process to a higher pressure. We use the ideal gas law and the reversible isothermal work formula to solve the problem.
The ideal gas law relates the pressure, volume, temperature, and number of moles of an ideal gas. We use it to find the number of moles of gas in the cylinder.
The reversible isothermal work formula relates the work done during a reversible isothermal process to the initial and final volumes of the gas. We use it to find the work required to compress the gas.
We use the natural logarithm function to find the ratio of final volume to initial volume, which is required in the work formula. We also use the logarithm table given in the question to find the natural logarithm of the ratio.
Finally, we get a negative value for the work required, which means work is done on the gas during the process. This is because the gas is being compressed, and work is required to compress a gas.
Given data:
Initial volume of the gas, V1 = 10 m3
Initial pressure of the gas, P1 = 2 bar
Final pressure of the gas, P2 = 15 bar
The process is isothermal, which means the temperature of the gas remains constant during the process.
To find the work done during the process, we can use the formula for reversible isothermal work:
W = nRT ln(V2/V1)
Where n is the number of moles of gas, R is the gas constant, T is the temperature of the gas, and V2/V1 is the ratio of the final volume to the initial volume.
Since the process is isothermal, the temperature remains constant, so we can use the ideal gas law to find the number of moles of gas:
PV = nRT
n = PV/RT
Where P and V are the initial pressure and volume of the gas, respectively, and R is the gas constant.
Substituting the values, we get:
n = (2 bar) x (10 m3) / (8.314 J/mol-K) x (273 K)
n = 0.9568 moles
Now we can calculate the final volume of the gas using the ideal gas law:
PV = nRT
V2 = nRT/P2
Substituting the values, we get:
V2 = (0.9568 moles) x (8.314 J/mol-K) x (273 K) / (15 bar)
V2 = 0.142 m3
So the ratio of final volume to initial volume is:
V2/V1 = 0.142 m3 / 10 m3
V2/V1 = 0.0142
Now we can calculate the work done during the process using the reversible isothermal work formula:
W = nRT ln(V2/V1)
Substituting the values, we get:
W = (0.9568 moles) x (8.314 J/mol-K) x (273 K) x ln(0.0142)
W = -6,652 J
So the work required for the process is -6,652 J, which means work is done on the gas during the process.
Explanation:
In this question, we are given the initial volume and pressure of an ideal gas and asked to find the work required to compress the gas in a reversible isothermal process to a higher pressure. We use the ideal gas law and the reversible isothermal work formula to solve the problem.
The ideal gas law relates the pressure, volume, temperature, and number of moles of an ideal gas. We use it to find the number of moles of gas in the cylinder.
The reversible isothermal work formula relates the work done during a reversible isothermal process to the initial and final volumes of the gas. We use it to find the work required to compress the gas.
We use the natural logarithm function to find the ratio of final volume to initial volume, which is required in the work formula. We also use the logarithm table given in the question to find the natural logarithm of the ratio.
Finally, we get a negative value for the work required, which means work is done on the gas during the process. This is because the gas is being compressed, and work is required to compress a gas.
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A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845?
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A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845?.
A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder contains 10 m3 of an ideal gas at a pressure of 2 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 15 bar. What quantum of work will be required for this process? (You can use the table given herewith.) Number 2 2.5 3 5 7 Log10 0.301 0.397 0.475 0.698 0.845?.
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