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  • In a given process of an ideal gas dW=0 and dQ<0. Then for the gasOption: 1 The temperature will decrease<st

In a given process of an ideal gas dW=0 and dQ<0. Then for the gas

Option: 1

The temperature will decrease


Option: 2

The pressure will increase


Option: 3

The volume will increase


Option: 4

The temperature will increase


Answers (1)

best_answer

As we learnt

 

Sign of dQ(Heat) -

 

If heat is imported or supplied to a system then dQ is positive. If heat is extracted from a system, then dQ is negative.

- wherein

dQ> 0 if heat is given

dQ< 0 if heat is extracted

 

So, from the question and by applying first law of thermodynamics

\Delta Q= \Delta U+\Delta W

Now,\\ \Delta W=0, \ \ \ So, \ \ \Delta Q = \Delta U \\ And \ \ \Delta Q<0, So,\ \ \Delta U <0,

 But for an ideal gas U\: \alpha \: T\Rightarrow dT<0

Hence temperature will decrease.

 

Posted by

Devendra Khairwa

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