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Diatomic gas undergoes result, the the adiabatic expansion against the piston of a cylinder. As a temperature of the gas drops from 1150 K to 400 K. The number of moles of gas required to obtain 2300 / of work from the expansion is constant R = 8.314 * 1mo * l ^ - 1 * K ^ - 1 (Round off to 2 decimal places) ( Аns :. 0 1475 )


can you explain the solution here how does efficiency becomes 7/5 & v =1.14?
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Diatomic gas undergoes result, the the adiabatic expansion against the...
Given information:
- The gas is diatomic.
- The temperature of the gas drops from 1150 K to 400 K.
- The gas undergoes an adiabatic expansion against the piston of a cylinder.
- The work done during the expansion is 2300 J.
- The gas constant is R = 8.314 J/(mol·K).

Calculating the change in volume:
To find the change in volume during the adiabatic expansion, we can use the formula:

ΔV = V2 - V1

Where:
- ΔV is the change in volume
- V2 is the final volume
- V1 is the initial volume

Since the number of moles of gas is constant, we can use the ideal gas law to relate the initial and final volumes to the initial and final temperatures:

V1/T1 = V2/T2

Where:
- V1 is the initial volume
- T1 is the initial temperature
- V2 is the final volume
- T2 is the final temperature

Rearranging the equation, we get:

V2 = V1 * (T2/T1)

Substituting the given values:
V2 = V1 * (400 K / 1150 K)

Calculating the number of moles of gas:
Next, we can use the ideal gas law to relate the number of moles of gas to the initial and final volumes:

n = (P1 * V1) / (R * T1)

Where:
- n is the number of moles of gas
- P1 is the initial pressure
- V1 is the initial volume
- R is the gas constant
- T1 is the initial temperature

Since the pressure is not given, we can assume it remains constant during the process, and therefore, P1 * V1 = P2 * V2. This allows us to cancel out the pressure term:

n = (V2 * P2) / (R * T1)

Since the number of moles of gas required to obtain 2300 J of work from the expansion is constant, we can set up the following equation:

n * R * (T2 - T1) = 2300 J

Substituting the values:
n * 8.314 J/(mol·K) * (400 K - 1150 K) = 2300 J

Simplifying the equation:
n * (-8.314 J/(mol·K) * 750 K) = 2300 J

Solving for n:
n = 2300 J / (-8.314 J/(mol·K) * 750 K)

Calculating n gives the number of moles of gas required to obtain the given work from the adiabatic expansion.
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Diatomic gas undergoes result, the the adiabatic expansion against the piston of a cylinder. As a temperature of the gas drops from 1150 K to 400 K. The number of moles of gas required to obtain 2300 / of work from the expansion is constant R = 8.314 * 1mo * l ^ - 1 * K ^ - 1 (Round off to 2 decimal places) ( Аns :. 0 1475 )can you explain the solution here how does efficiency becomes 7/5 & v =1.14?
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Diatomic gas undergoes result, the the adiabatic expansion against the piston of a cylinder. As a temperature of the gas drops from 1150 K to 400 K. The number of moles of gas required to obtain 2300 / of work from the expansion is constant R = 8.314 * 1mo * l ^ - 1 * K ^ - 1 (Round off to 2 decimal places) ( Аns :. 0 1475 )can you explain the solution here how does efficiency becomes 7/5 & v =1.14? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Diatomic gas undergoes result, the the adiabatic expansion against the piston of a cylinder. As a temperature of the gas drops from 1150 K to 400 K. The number of moles of gas required to obtain 2300 / of work from the expansion is constant R = 8.314 * 1mo * l ^ - 1 * K ^ - 1 (Round off to 2 decimal places) ( Аns :. 0 1475 )can you explain the solution here how does efficiency becomes 7/5 & v =1.14? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Diatomic gas undergoes result, the the adiabatic expansion against the piston of a cylinder. As a temperature of the gas drops from 1150 K to 400 K. The number of moles of gas required to obtain 2300 / of work from the expansion is constant R = 8.314 * 1mo * l ^ - 1 * K ^ - 1 (Round off to 2 decimal places) ( Аns :. 0 1475 )can you explain the solution here how does efficiency becomes 7/5 & v =1.14?.
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