Question

Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be
(a) 500 J
(b) 460 J
(c) 300 J
(d) 380 J

Answer 1

in path $A\rightarrow B$ :

it is isochoric path, and we know, in isochoric process workdone will be zero.

e.g., $W_{A\rightarrow B}=0$

so, $\Delta Q_{A\rightarrow B}=\Delta U_{A\rightarrow B}$ = 400J

in path $B\rightarrow C$ :

it is isobaric process. so workdone will be $W_{B\rightarrow C}=P\Delta V_{B\rightarrow C}=P(V_C-V_B)$

so, $\Delta Q_{B\rightarrow C}=\Delta U_{B\rightarrow C}+P(V_C-V_B)$

100 J = $\Delta U_{B\rightarrow C}$ + 6 × 10⁴(4 × 10^-3 - 2 × 10^-3)

$\Delta U_{B\rightarrow}$ = -20J

in path, $C\rightarrow A$

$\Delta Q_{A\rightarrow C}=\Delta U_{A\rightarrow C}+\textbf{area of triangle made by PV graph}$

= $\Delta U_{A\rightarrow B}+\Delta U_{B\rightarrow C}+\frac{1}{2}(6 -2) × 10^4 × (4 - 2) × 10^3$

= $400-20+80$

= 460 J

hence, option (b) is correct choice.

Answer 2

Explanation:

ANSWER

In process A to B change in internal energy is

ΔU

AB

=ΔQ

AB

−ΔW

AB

process AB is Isochoric, therefore heat added is equal to change in internal energy.

ΔU

AB

=ΔQ

AB

=400J

process BC change in internal energy is

ΔU

BC

=ΔQ

BC

−ΔW

BC

ΔU

BC

=100−PΔV

ΔU

BC

=100−6×10

4

(4×10

−3

−2×10

−3

)

ΔU

BC

=100−6×10

4

×2×10

−3

ΔU

BC

=100−120J=−20J

ΔU

BC

=100−6×10

4

×2×10

−3

total change in internal energy will be :

ΔU

AC

=400−20=380J

heat added in process AC will be:

ΔQ

AC

=ΔU

AC

+ΔW

AC

ΔQ

AC

=380+areaofP−Vdiagram

ΔQ

AC

=380+

2

1

(6×10

4

+2×10

4

)2×10

−3

ΔQ

AC

=380+80=460J