An ideal gas undergoes the cyclic process abca as shown in the figure. The internal energy change of the gas along the path ca is -180 J. The gas absorb 250 J of heat along the path ab and also absorbs 60 J of heat along the path bc Find the work done by the gas along the path abc

Question

An ideal gas undergoes the cyclic process abca as shown in the figure. The internal energy change of the gas along the path ca is -180 J. The gas absorb 250 J of heat along the path ab and also absorbs 60 J of heat along the path bc Find the work done by the gas along the path abc <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jbrb4mnbqnmd3zia.jpg" /> (A) 130 J (B) 140 J (C) 120 J (D) 100 J

Tardigrade Admin · Accepted Answer

Change in internal energy for a cyclic process is zero. Thus, Δ Uabc+Δ Uca=0Δ Uabc=-Δ Uca=180 J ΔQ=ΔW+ΔU 250+60=ΔW+180 ΔW=130 J

$130 J$

$140 J$

$120 J$

$100 J$

Change in internal energy for a cyclic process is zero. Thus,
$Δ U_{ab c} + Δ U_{c a} = 0Δ U_{ab c} = - Δ U_{c a} = 180 J$
$Δ Q = Δ W + Δ U$
$250 + 60 = Δ W + 180$
$Δ W = 130 J$

Q. An ideal gas undergoes the cyclic process abca as shown in the figure. The internal energy change of the gas along the path ca is −180J. The gas absorb 250J of heat along the path ab and also absorbs 60J of heat along the path bc Find the work done by the gas along the path abc

130J

140J

120J

100J

Change in internal energy for a cyclic process is zero. Thus, ΔUabc​+ΔUca​=0ΔUabc​=−ΔUca​=180J ΔQ=ΔW+ΔU 250+60=ΔW+180 ΔW=130J

$130 J$

$140 J$

$120 J$

$100 J$

Change in internal energy for a cyclic process is zero. Thus,
$Δ U_{ab c} + Δ U_{c a} = 0Δ U_{ab c} = - Δ U_{c a} = 180 J$
$Δ Q = Δ W + Δ U$
$250 + 60 = Δ W + 180$
$Δ W = 130 J$