In a given process of an ideal gas, ∆W = 0 and ∆Q=-ve. Then for the gas:
Correct answer: A. The temperature will decrease
- A. The temperature will decrease
- B. The volume will increase
- C. The pressure will remain constant
- D. The temperature will increase
Explanation
If ∆W = 0 and ∆Q = -ve for an ideal gas, then the temperature of the gas will decrease. This is because the first law of thermodynamics states that ∆U = ∆Q - ∆W, where ∆U is the change in internal energy of the gas, ∆Q is the heat added to the gas, and ∆W is the work done by the gas. If ∆W= 0 and ∆Q = -ve, then ∆U < 0, which means that the internal energy of the gas decreases. Since the internal energy of the gas is decreasing, the temperature of the gas must also decrease.
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About First Law of Thermodynamics
The first law relates heat supplied, work done and the change in internal energy through energy conservation. Problems use sign conventions and apply the law to isothermal, adiabatic, isobaric and isochoric processes. Internal energy is a state function, while heat and work depend on the path followed.
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