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Entropy for a Reversible Process

For an ideal ...

Question

For an ideal gas, an illustration of three different paths $A$ , $(B + C)$ and $(D + E)$ from an initial state $P_{1}$ , $V_{1}$ , $T_{1}$ to a final state $P_{2}$ , $V_{2}$ , $T_{1}$ is shown in the given figure.
Path A represents a reversible Isothermal expansion from
$P_{1}$ , $V_{1} t o P_{2}, V_{2}$ . Path $(B + C)$ reperesents a reversible adiabatic expansion
$(B)$ from $P_{1} V_{1}$ , $T_{1} t o P_{3}$ , $V_{2}$ , $T_{2}$ followed by reversible heating the gas at constant volume (C) from $P_{3}$ , $V_{2}$ , $T_{2} t o P_{2}$ , $V_{2}$ , $T_{1}$ . Path ?
$(D + E)$ represents a reversible expansion at constant pressure $P_{1} (D)$
from $P_{1}$ , $V_{1}$ , $T_{1}$ to $P_{1}$ , $V_{2}$ , $T_{3}$ followed by a reversible cooling at constant volume $V_{2} (E)$
from $P_{1}$ , $V_{2}$ , $T_{3} t o P_{2}$ , $V_{2}$ , $T_{1}$
What is $q_{r e v}$ , for path $(A)$ ?

P (V_{2} - V_{1})

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- n R T_{3} l n \frac{V_{2}}{V_{1}}

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- n R l n \frac{V_{2}}{V_{1}}

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- n R T_{1} l n \frac{V_{2}}{V_{1}}

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Solution

The correct option is D $- n R T_{1} l n \frac{V_{2}}{V_{1}}$
According to the first law of Thermodynamics, $Δ U = q - w$ in an isothermal process.
$Δ U = 0$ , hence
$q = w$ or $w = n R T_{1} l n \frac{V_{2}}{V_{1}}$

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Similar questions

Q. For an ideal gas, an illustration of three different paths

A

(B + C)

and

(D + E)

from an initial state

P_{1}

V_{1}

T_{1}

to a final state

P_{2}

V_{2}

T_{1}

is shown in the given figure:

Path A represents a reversible Isothermal expansion from

P_{1}

V_{1} t o P_{2}, V_{2}

. Path

(B + C)

reperesents a reversible adiabatic expansion

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from

P_{1} V_{1}

T_{1} t o P_{3}

V_{2}

T_{2}

followed by reversible heating the gas at constant volume (C) from

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T_{2} t o P_{2}

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T_{1}

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represents a reversible expansion at constant pressure

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from

P_{1}

V_{1}

T_{1}

P_{1}

V_{2}

T_{3}

followed by a reversible cooling at constant volume

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from

P_{1}

V_{2}

T_{3} t o P_{2}

V_{2}

T_{1}

What is

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(B + C) ?

For an ideal gas, an illustration of three different paths

A

(B + C)

and

(D + E)

from an initial state

P_{1}

V_{1}

T_{1}

to a final state

P_{2}

V_{2}

T_{1}

is shown in the given figure.
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P_{1}

V_{1}

P_{2}, V_{2}

. Path

(B + C)

reperesents a reversible adiabatic expansion

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P_{1} V_{1}

T_{1}

P_{3}

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followed by reversible heating the gas at constant volume

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P_{3}

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P_{2}

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V_{2}

T_{3}

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V_{2} (E)

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The correct option is D −nRT1lnV2V1According to the first law of Thermodynamics, ΔU=q−w in an isothermal process. ΔU=0, hence q=w or w=nRT1lnV2V1

The correct option is D $- n R T_{1} l n \frac{V_{2}}{V_{1}}$
According to the first law of Thermodynamics, $Δ U = q - w$ in an isothermal process.
$Δ U = 0$ , hence
$q = w$ or $w = n R T_{1} l n \frac{V_{2}}{V_{1}}$