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 work is done by the system: $Δ U = q + w$
When no work is done: $△ U = q$
In gaseous system: $Δ U = q + P Δ V$

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?
At constant temperature: $q = - w$
When work is done by the system: $Δ U = q + w$
When no work is done: $△ U = q$
In gaseous system: $Δ U = q + P Δ V$

Toppr Admin · Accepted Answer

At constant temperature- -x394-T-0-x2234-x394-U-nCv-x394-T-x2234-x394-U-q-w-0q-x2212-wWhen work done is zero- w-0-x2234-x394-U-q-w-x2234-x394-U-qIn gaseous system- w-P-x394-V-x2234-x394-U-q-w-x2234-x394-U-q-P-x394-VWhen work is done by the system- w is negative-x2234-x394-U-q-x2212-w

At constant temperature, ΔT=0∴ΔU=nCvΔT∴ΔU=q+w=0q=−wWhen work done is zero, w=0∴ΔU=q+w∴ΔU=qIn gaseous system, w=PΔV∴ΔU=q+w∴ΔU=q+PΔVWhen work is done by the system, w is negative.∴ΔU=q−w

At constant temperature, $Δ T = 0$
$∴ Δ U = n C_{v} Δ T ∴ Δ U = q + w = 0 q = - w$
When work done is zero, $w = 0$
$∴ Δ U = q + w ∴ Δ U = q$
In gaseous system, $w = P Δ V$
$∴ Δ U = q + w ∴ Δ U = q + P Δ V$
When work is done by the system, w is negative.
$∴ Δ U = q - w$