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For a particular thermodynamics system the entropy U and volume V by S=cu^(14)y^(14) where c is a constant.The Gibbs potential G=U-TS p1 for this system is (a) (3p)/(47) (b) (cl)/(3) (c) zero (d) (omega)/(47))?
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For a particular thermodynamics system the entropy U and volume V by S...

Explanation:

Given:
- Entropy, S = cu^(14)y^(14)
- Volume, V = yu

Calculating Gibbs Potential G:
- Gibbs potential, G = U - TS
- Substituting the given values, G = cu^(14)y^(14) - Tcu^(14)y^(14)
- G = cu^(14)y^(14)(1 - T)

Finding Gibbs Potential at equilibrium (G=0):
- At equilibrium, G = 0
- Therefore, cu^(14)y^(14)(1 - T) = 0
- Since cu^(14)y^(14) cannot be zero, we have 1 - T = 0
- T = 1

Substitute T=1 into G:
- G = cu^(14)y^(14)(1 - 1)
- G = 0

Conclusion:
- The Gibbs potential G for this system at equilibrium is zero.
- Therefore, the correct option is (c) zero.
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For a particular thermodynamics system the entropy U and volume V by S=cu^(14)y^(14) where c is a constant.The Gibbs potential G=U-TS p1 for this system is (a) (3p)/(47) (b) (cl)/(3) (c) zero (d) (omega)/(47))?
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For a particular thermodynamics system the entropy U and volume V by S=cu^(14)y^(14) where c is a constant.The Gibbs potential G=U-TS p1 for this system is (a) (3p)/(47) (b) (cl)/(3) (c) zero (d) (omega)/(47))? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about For a particular thermodynamics system the entropy U and volume V by S=cu^(14)y^(14) where c is a constant.The Gibbs potential G=U-TS p1 for this system is (a) (3p)/(47) (b) (cl)/(3) (c) zero (d) (omega)/(47))? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a particular thermodynamics system the entropy U and volume V by S=cu^(14)y^(14) where c is a constant.The Gibbs potential G=U-TS p1 for this system is (a) (3p)/(47) (b) (cl)/(3) (c) zero (d) (omega)/(47))?.
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