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One mole of a...

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One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by $P V^{n} = C o n s t a n t$ . We can conclude that $(T = t e m p e r a t u r e, W = w o r k$ done by gas, $V = v o l u m e$ and $U = i n t e r n a l e n e r g y)$

A

T_{A} = T_{B} = T_{C}

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B

V_{A} < V_{B}, P_{B} < P_{C}

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C

W_{A B} > W_{B C}

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D

U_{A} < U_{B}

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Solution

The correct option is C $W_{A B} > W_{B C}$
$Solution :$
By PV = nRT
$T_{B} > T_{A}$
$W_{A B} = P_{0} (2 V_{0} - V_{0}) = P_{0} V_{0}$
$W_{B C} = R T 1 n \frac{3}{2} = 2 P_{0} V_{0} [1 n 3 - 1 n 2] = 2 P_{0} V_{0} (2.3030) (0.477 - 0.30) = 0.8142 P_{0} V_{0} < W_{A B}$

$Hence C is the correct option$

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Q. One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by

P V^{n}

= constant. We can conclude that (where T = temperature, W = work done by gas, V = volume and U = internal energy).

Q. One mole of a monoatomic ideal gas is taken through the cycle shown in figure.

A \to B :

Adiabatic expansion

B \to C :

Cooling at constant volume

C \to D :

Adiabatic compression.

D \to A :

Heating at constant volume
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(P_{A}, T_{A}), (P_{B}, T_{B}), (P_{C}, T_{C})

(P_{D}, T_{D})

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moles of an ideal monoatomic gas performs

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Q. One mole of a diatomic ideal gas

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\frac{V_{A}}{V_{B}} = 16, \frac{V_{C}}{V_{B}} = 2

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W_{A B} = - 1522.5 R J, W_{B C} = 909 R J, W_{C D} = 2567.5 R J

One mole of a monoatomic ideal gas is taken through the cycle as shown in figure:

A \to B

: adiabatic expansion

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: cooling at constant volume

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P_{A}, T_{A}, P_{B}, T_{B}

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T_{A} = 1000 K

P_{B} = \frac{2}{3} P_{A}

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[\frac{2}{3}]^{2 / 5} = 0.85

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