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A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process until its pressure increases to 5 bar.
The work in kJ required for this process is _____ (take ln 5= 1.609)
- a)804.7
- b)953.2
- c)981.7
- d)1012.2
Correct answer is option 'A'. Can you explain this answer?
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A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This...
Work done for an isothermal process
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The work required is 804.72 kJ.
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A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This...
Given:
- Volume of gas (V1) = 5 m^3
- Initial pressure of gas (P1) = 1 bar
- Final pressure of gas (P2) = 5 bar
- Natural logarithm of 5 (ln 5) = 1.609
To find:
- Work done in the process (W)
Assumption:
- The gas is assumed to be an ideal gas.
- The process is assumed to be reversible and isothermal.
Explanation:
1. Isothermal Process:
In an isothermal process, the temperature of the gas remains constant. This means that the product of pressure and volume (PV) of the gas remains constant throughout the process.
P1V1 = P2V2
2. Calculating Final Volume (V2):
Using the given values, we can rearrange the equation to find the final volume (V2).
V2 = (P1V1) / P2
V2 = (1 * 5) / 5
V2 = 1 m^3
3. Work done (W):
The work done in an isothermal process can be calculated using the following equation:
W = nRT * ln(V2 / V1)
Where:
- n = number of moles of gas
- R = ideal gas constant
- T = temperature of the gas (constant in this case)
- ln = natural logarithm function
4. Calculating Number of Moles (n):
To calculate the number of moles, we need to use the ideal gas equation:
PV = nRT
Rearranging the equation, we get:
n = PV / RT
5. Calculating Work done (W):
Substituting the values into the equation for work done:
W = nRT * ln(V2 / V1)
W = (PV / RT) * RT * ln(V2 / V1)
W = P * V * ln(V2 / V1)
W = 1 * 5 * ln(1 / 5)
W = 5 * ln(0.2)
W = 5 * (-1.609)
W = -8.045 kJ
6. Final Answer:
Since work done is a scalar quantity, the negative sign indicates that work is done on the gas. Taking the absolute value, the work done in this process is approximately 8.045 kJ.
The correct answer is option A) 804.7.
- Volume of gas (V1) = 5 m^3
- Initial pressure of gas (P1) = 1 bar
- Final pressure of gas (P2) = 5 bar
- Natural logarithm of 5 (ln 5) = 1.609
To find:
- Work done in the process (W)
Assumption:
- The gas is assumed to be an ideal gas.
- The process is assumed to be reversible and isothermal.
Explanation:
1. Isothermal Process:
In an isothermal process, the temperature of the gas remains constant. This means that the product of pressure and volume (PV) of the gas remains constant throughout the process.
P1V1 = P2V2
2. Calculating Final Volume (V2):
Using the given values, we can rearrange the equation to find the final volume (V2).
V2 = (P1V1) / P2
V2 = (1 * 5) / 5
V2 = 1 m^3
3. Work done (W):
The work done in an isothermal process can be calculated using the following equation:
W = nRT * ln(V2 / V1)
Where:
- n = number of moles of gas
- R = ideal gas constant
- T = temperature of the gas (constant in this case)
- ln = natural logarithm function
4. Calculating Number of Moles (n):
To calculate the number of moles, we need to use the ideal gas equation:
PV = nRT
Rearranging the equation, we get:
n = PV / RT
5. Calculating Work done (W):
Substituting the values into the equation for work done:
W = nRT * ln(V2 / V1)
W = (PV / RT) * RT * ln(V2 / V1)
W = P * V * ln(V2 / V1)
W = 1 * 5 * ln(1 / 5)
W = 5 * ln(0.2)
W = 5 * (-1.609)
W = -8.045 kJ
6. Final Answer:
Since work done is a scalar quantity, the negative sign indicates that work is done on the gas. Taking the absolute value, the work done in this process is approximately 8.045 kJ.
The correct answer is option A) 804.7.
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A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process until its pressure increases to 5 bar.The work in kJ required for this process is _____ (take ln 5= 1.609)a)804.7b)953.2c)981.7d)1012.2Correct answer is option 'A'. Can you explain this answer?
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