An ideal gas undergoes a cyclic process as shown in Figure. Δ UBC = - 5 kJ mol-1 , qAB = 2 kJ mol-1 WAB = - 5 kJ mol-1 , WCA = 3 kJ mol-1 Heat absorbed by the system during process CA is :

Question

An ideal gas undergoes a cyclic process as shown in Figure. Δ UBC = - 5 kJ mol-1 , qAB = 2 kJ mol-1 WAB = - 5 kJ mol-1 , WCA = 3 kJ mol-1 Heat absorbed by the system during process CA is : (A) -5 kJ mol-1 (B)  + 5 kJ mol-1 (C) 18 kJ mol-1 (D)  - 18 kJ mol-1

Tardigrade Admin · Accepted Answer

We have Δ U AB =q AB +w AB =2+(-5)=-3 kJ mol -1 For a cyclic process, Δ U=0 . Therefore, Δ U=Δ UA B+Δ UB C+Δ UC A Δ U CA =-Δ U AB -Δ U BC =-(-3)-(-5)=8 kJ mol -1 Δ U CA =q CA +w CA 8 =q CA +3 ⇒ q CA =5 kJ mol -1

Q. An ideal gas undergoes a cyclic process as shown in Figure.
$Δ U_{BC} = - 5 k J m o l^{- 1}, q_{A B} = 2 k J m o l^{- 1}$
$W_{A B} = - 5 k J m o l^{- 1}, W_{C A} = 3 k J m o l^{- 1}$
Heat absorbed by the system during process $C A$ is :

$- 5 k J m o l^{- 1}$

$+ 5 k J m o l^{- 1}$

$18 k J m o l^{- 1}$

$- 18 k J m o l^{- 1}$

Q. An ideal gas undergoes a cyclic process as shown in Figure. ΔUBC​=−5kJmol−1,qAB​=2kJmol−1 WAB​=−5kJmol−1,WCA​=3kJmol−1 Heat absorbed by the system during process CA is :

−5kJmol−1

+5kJmol−1

18kJmol−1

−18kJmol−1

We have ΔUAB​=qAB​+wAB​ =2+(−5)=−3kJmol−1 For a cyclic process, ΔU=0. Therefore, ΔU=ΔUAB​+ΔUBC​+ΔUCA​ ΔUCA​=−ΔUAB​−ΔUBC​ =−(−3)−(−5)=8kJmol−1 ΔUCA​=qCA​+wCA​ 8=qCA​+3⇒qCA​=5kJmol−1

Q. An ideal gas undergoes a cyclic process as shown in Figure.
$Δ U_{BC} = - 5 k J m o l^{- 1}, q_{A B} = 2 k J m o l^{- 1}$
$W_{A B} = - 5 k J m o l^{- 1}, W_{C A} = 3 k J m o l^{- 1}$
Heat absorbed by the system during process $C A$ is :

$- 5 k J m o l^{- 1}$

$+ 5 k J m o l^{- 1}$

$18 k J m o l^{- 1}$

$- 18 k J m o l^{- 1}$