During the thermodynamic process shown in figure for an ideal gas

Question

During the thermodynamic process shown in figure for an ideal gas (A) Δ T=0 (B) Δ Q=0 (C) W < 0 (D) Δ U > 0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/01def64a9f1a52ccba1453d7ec2acf9c-.png" /> For a straight P-V graph line P propto V If pressure increases, volume increases then T also increases [P V propto T] So Δ T ≠ 0 Volume increasing so work is positive, W>0 and temperature also increasing so Δ Q>0 ∵ Δ Q=Δ U+Δ W So Δ U>0

Q. During the thermodynamic process shown in figure for an ideal gas

$Δ T = 0$

$Δ Q = 0$

$W < 0$

$Δ U > 0$

For a straight $P - V$ graph line $P \propto V$
If pressure increases, volume increases then $T$ also increases $[P V \propto T]$
So $Δ T \neq = 0$
Volume increasing so work is positive, $W > 0$
and temperature also increasing so $Δ Q > 0$
$∵ Δ Q = Δ U + Δ W$
So $Δ U > 0$

Q. During the thermodynamic process shown in figure for an ideal gas

ΔT=0

ΔQ=0

W<0

ΔU>0

For a straight P−V graph line P∝V If pressure increases, volume increases then T also increases [PV∝T] So ΔT=0 Volume increasing so work is positive, W>0 and temperature also increasing so ΔQ>0 ∵ΔQ=ΔU+ΔW So ΔU>0

$Δ T = 0$

$Δ Q = 0$

$W < 0$

$Δ U > 0$

For a straight $P - V$ graph line $P \propto V$
If pressure increases, volume increases then $T$ also increases $[P V \propto T]$
So $Δ T \neq = 0$
Volume increasing so work is positive, $W > 0$
and temperature also increasing so $Δ Q > 0$
$∵ Δ Q = Δ U + Δ W$
So $Δ U > 0$