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A monoatomic ideal gas expands isobarically;the percentage of heat supplies that increases internal energy and that involved in doing work for expansion is (respectively)??answer is 60:40;;;;EXPLAIN. .?
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A monoatomic ideal gas expands isobarically;the percentage of heat sup...
Isobaric Expansion of a Monoatomic Ideal Gas

Introduction:
In an isobaric process, the pressure of the system remains constant while the volume changes. When a monoatomic ideal gas undergoes an isobaric expansion, heat is supplied to the system, which increases the internal energy of the gas and enables it to perform work to expand against the external pressure.

Percentage of Heat Supplied:
The percentage of heat supplied that increases the internal energy of the gas can be determined using the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat supplied (Q) minus the work done (W) by the system:

ΔU = Q - W

Since the process is isobaric, the work done can be calculated as:

W = PΔV

Where P is the constant pressure and ΔV is the change in volume. Therefore, the equation becomes:

ΔU = Q - PΔV

To calculate the percentage of heat supplied that increases the internal energy, we can rearrange the equation:

(Q - ΔU) / Q = PΔV / Q

Since the process is isobaric, ΔV / Q can be expressed as the change in volume per unit of heat supplied, which is also known as the coefficient of volume expansion (β):

(Q - ΔU) / Q = Pβ

The coefficient of volume expansion (β) depends on the properties of the gas and remains constant for a given gas. Therefore, the percentage of heat supplied that increases the internal energy is directly proportional to the pressure (P) and the coefficient of volume expansion (β).

Percentage of Heat Involved in Doing Work for Expansion:
The remaining percentage of heat supplied is involved in doing work for the expansion of the gas. Using the equation:

(Q - ΔU) / Q = Pβ

We can rearrange the equation to find the percentage of heat involved in doing work:

ΔU / Q = 1 - Pβ

Therefore, the percentage of heat involved in doing work for expansion is equal to 1 minus the percentage of heat supplied that increases the internal energy.

Conclusion:
In an isobaric expansion of a monoatomic ideal gas, the percentage of heat supplied that increases the internal energy is directly proportional to the pressure and the coefficient of volume expansion. The remaining percentage of heat is involved in doing work for the expansion. The specific values of the percentages may vary depending on the specific conditions of the system, but the ratio of 60:40 (heat supplied:work done) is a commonly observed proportion in many isobaric processes.
Community Answer
A monoatomic ideal gas expands isobarically;the percentage of heat sup...
Cp for monoatomic is 5R/2 and Cv is 3R/2
so part of heat in increasing internal energy will be ∆U/∆Q = n×Cv×∆T/ n×Cp×∆T = Cv/Cp=1.5R/2.5R ×100=60%
and part of heat in work expansion = W/∆Q = n×R×∆T/n×Cp×∆T = R/2.5R ×100% =40% so ratio of parts of heat supplied to both is 60/40.
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A monoatomic ideal gas expands isobarically;the percentage of heat supplies that increases internal energy and that involved in doing work for expansion is (respectively)??answer is 60:40;;;;EXPLAIN. .?
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