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The ratio of slopes of log P vs log V for reversible adiabatic process and reversible isothermal process of an ideal gas is equal to :
- a)γ
- b)1-γ
- c)γ-1
- d)1/γ
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The ratio of slopes of log P vs log V for reversible adiabatic process...
PVγ= constant for adiabatic expansion
and PV = constant for isothermal expansion
∴ log P =-γlogγ slope = -γ
log P =-logV slope = -1
and PV = constant for isothermal expansion
∴ log P =-γlogγ slope = -γ
log P =-logV slope = -1
Most Upvoted Answer
The ratio of slopes of log P vs log V for reversible adiabatic process...
1
Explanation:
For a reversible adiabatic process, we have:
PV^γ = constant, where γ = Cp/Cv
Taking logarithm on both sides, we get:
log P + γ log V = constant
Differentiating with respect to log V, we get:
(∂log P/∂log V)adiabatic = -γ
For a reversible isothermal process, we have:
PV = constant
Taking logarithm on both sides, we get:
log P + log V = constant
Differentiating with respect to log V, we get:
(∂log P/∂log V)isothermal = -1
Therefore, the ratio of slopes of log P vs log V for reversible adiabatic process and reversible isothermal process is:
(∂log P/∂log V)adiabatic / (∂log P/∂log V)isothermal = γ/1 = γ
Since for an ideal gas, γ = Cp/Cv = 1 + 2/f, where f is the degree of freedom, the ratio of slopes depends on the degree of freedom of the gas. For a monoatomic gas (f = 3), γ = 5/3, and for a diatomic gas (f = 5), γ = 7/5.
Explanation:
For a reversible adiabatic process, we have:
PV^γ = constant, where γ = Cp/Cv
Taking logarithm on both sides, we get:
log P + γ log V = constant
Differentiating with respect to log V, we get:
(∂log P/∂log V)adiabatic = -γ
For a reversible isothermal process, we have:
PV = constant
Taking logarithm on both sides, we get:
log P + log V = constant
Differentiating with respect to log V, we get:
(∂log P/∂log V)isothermal = -1
Therefore, the ratio of slopes of log P vs log V for reversible adiabatic process and reversible isothermal process is:
(∂log P/∂log V)adiabatic / (∂log P/∂log V)isothermal = γ/1 = γ
Since for an ideal gas, γ = Cp/Cv = 1 + 2/f, where f is the degree of freedom, the ratio of slopes depends on the degree of freedom of the gas. For a monoatomic gas (f = 3), γ = 5/3, and for a diatomic gas (f = 5), γ = 7/5.
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The ratio of slopes of log P vs log V for reversible adiabatic process and reversible isothermal process of an ideal gas is equal to :a)γb)1-γc)γ-1d)1/γCorrect answer is option 'A'. Can you explain this answer?
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