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One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7?
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One mole of an ideal monoatomic gas is compressed isothermally in a ri...
Work done in an isothermal process is
nRTln (v2÷v1). just get your answer from this equation.
nRTln (v2÷v1). just get your answer from this equation.
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One mole of an ideal monoatomic gas is compressed isothermally in a ri...
Problem:
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be:
(1) 300 R
(2) 300 R ln 6
(3) 300 R ln 2
(4) 300 R ln 7
Solution:
Given:
- One mole of an ideal monoatomic gas
- Isothermal compression
- Pressure doubles
- Room temperature, 27°C
Formula:
The work done on a gas during an isothermal process is given by the equation:
W = -nRT ln(Vf/Vi)
Where:
W = work done on the gas
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Vf = final volume
Vi = initial volume
Step 1: Convert temperature to Kelvin
Given temperature = 27°C
Convert to Kelvin: 27 + 273 = 300 K
Step 2: Determine the initial and final volumes
Since the gas is compressed isothermally, the temperature remains constant. Therefore, the initial and final volumes of the gas are inversely proportional to the initial and final pressures.
If the initial pressure is P, then the final pressure is 2P.
Using the ideal gas law, PV = nRT, we can calculate the initial and final volumes:
Vi = nRT/P
Vf = nRT/(2P)
Step 3: Substitute the values into the work equation
W = -nRT ln(Vf/Vi)
= -nRT ln((nRT/(2P))/(nRT/P))
= -nRT ln((nRT)/(2nRT))
= -nRT ln(1/2)
= nRT ln(2)
Step 4: Calculate the work done
W = nRT ln(2)
= (1 mol)(R)(300 K) ln(2)
= 300 R ln(2)
Answer:
The work done on the gas during the isothermal compression is 300 R ln(2). Therefore, the correct option is (3) 300 R ln(2).
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be:
(1) 300 R
(2) 300 R ln 6
(3) 300 R ln 2
(4) 300 R ln 7
Solution:
Given:
- One mole of an ideal monoatomic gas
- Isothermal compression
- Pressure doubles
- Room temperature, 27°C
Formula:
The work done on a gas during an isothermal process is given by the equation:
W = -nRT ln(Vf/Vi)
Where:
W = work done on the gas
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Vf = final volume
Vi = initial volume
Step 1: Convert temperature to Kelvin
Given temperature = 27°C
Convert to Kelvin: 27 + 273 = 300 K
Step 2: Determine the initial and final volumes
Since the gas is compressed isothermally, the temperature remains constant. Therefore, the initial and final volumes of the gas are inversely proportional to the initial and final pressures.
If the initial pressure is P, then the final pressure is 2P.
Using the ideal gas law, PV = nRT, we can calculate the initial and final volumes:
Vi = nRT/P
Vf = nRT/(2P)
Step 3: Substitute the values into the work equation
W = -nRT ln(Vf/Vi)
= -nRT ln((nRT/(2P))/(nRT/P))
= -nRT ln((nRT)/(2nRT))
= -nRT ln(1/2)
= nRT ln(2)
Step 4: Calculate the work done
W = nRT ln(2)
= (1 mol)(R)(300 K) ln(2)
= 300 R ln(2)
Answer:
The work done on the gas during the isothermal compression is 300 R ln(2). Therefore, the correct option is (3) 300 R ln(2).
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One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7?
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One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7?.
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27C. The work done on the gas will be : (1) 300 R (2) 300 R ln 6 (3) 300 R ln 2 (4) 300 R ln 7?.
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