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One mole of an ideal gas is taken from state A to state B b

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One mole of an ideal gas is taken from state $A$ to state $B$ by three different processes (a) $A CB,$ (b) $A D B$ and(c) $A EB$ as shown in the $P - V$ diagram. The heat absorbed by the gas is

greater in process (b) than in (a)

the least in process (b)

the same in (a) and (c)

less in (c) than in (b)

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Benjamin Discussed

One mole of an ideal gas is taken from state

A

to state

B

by three different processes (a)

A CB,

(b)

A D B

and(c)

A EB

as shown in the

P - V

diagram. The heat absorbed by the gas is

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Heat absorbed by gas in three processes is given by

Q_{A CB} = Δ U + W_{A CB}

Q_{A D B} = Δ U Q_{A EB} = Δ U + W_{A EB}

The change in internal energy in all the three cases is same and

W_{A CB}

is positive,

W_{A EB}

is negative.

Hence

Q_{A CB} > Q_{A D B} > Q_{A EB}

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Question Text	One mole of an ideal gas is taken from state $A$ to state $B$ by three different processes (a) $A CB,$ (b) $A D B$ and(c) $A EB$ as shown in the $P - V$ diagram. The heat absorbed by the gas is
Answer Type	Text solution:1
Upvotes	150

One mole of an ideal gas is taken from state A to state B by three different processes (a) ACB, (b) ADB and(c) AEB as shown in the P−V diagram. The heat absorbed by the gas is

Text solutionVerified

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One mole of an ideal gas is taken from state $A$ to state $B$ by three different processes (a) $A CB,$ (b) $A D B$ and(c) $A EB$ as shown in the $P - V$ diagram. The heat absorbed by the gas is