In a given process of an ideal gas, dW = 0 and dQ

Question

In a given process of an ideal gas, dW = 0 and dQ < 0. The for the gas
Select one:

a)
the volume will increase
b)
the temperature will decrease
c)
the pressure will remain constant
d)
the temperature will increase

Correct answer is option 'B'. Can you explain this answer?

Answer

From first law of thermodynamics,
dQ = dU + dW
dQ = dU if dW = 0
Since, dQ < 0
Therefore,

or temperature will decrease
The correct answer is: the temperature will decrease

Answer

From first law of thermodynamics,
dQ = dU + dW
dQ = dU if dW = 0
Since, dQ < 0
Therefore,

or temperature will decrease
The correct answer is: the temperature will decrease

Answer

= 0. This means that no work is done on or by the gas and no heat is transferred to or from the gas during the process.

In such a process, the internal energy of the gas remains constant. This can be expressed mathematically as:

ΔU = Q - W = 0

Since dQ = 0, this implies that ΔU = 0, which means that the internal energy of the gas does not change.

Overall, this type of process is called an isochoric process or an isovolumetric process, where the volume of the gas remains constant.

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