According to the first law of thermodynamics, Δ U = q + w. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression?

Question

According to the first law of thermodynamics, Δ U = q + w. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression? (A) At constant temperature: q - -w (B) When no work is done: Δ U = q (C) In gaseous system: Δ U = q + PΔ V (D) When work is done by the system: Δ U = q + w

Tardigrade Admin · Accepted Answer

When work is done by the system, Δ U=q-w.

Q. According to the first law of thermodynamics, $Δ U = q + w$ . In special cases the statement can be expressed in different ways. Which of the following is not a correct expression?

At constant temperature : $q - - w$

When no work is done : $Δ U = q$

In gaseous system : $Δ U = q + P Δ V$

When work is done by the system : $Δ U = q + w$

When work is done by the system, $Δ U = q - w$ .

Q. According to the first law of thermodynamics, ΔU=q+w. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression?

At constant temperature : q−−w

When no work is done : ΔU=q

In gaseous system : ΔU=q+PΔV

When work is done by the system : ΔU=q+w