Question 1: Thermodynamics & KTG

Question

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at
pressure P
i
= 10
5
Pa and volume V
i
= 10
-3
m
3
changes to a final state at P
f
= $\left( {{1 \over {32}}} ight) 	imes {10^5}\,Pa$ and
V
f
= 8 $ 	imes $ 10
-3
m
3
in an adiabatic quasi-static process, such that P
3
V
5
= constant. Consider another
thermodynamic process that brings the system from the same initial state to the same final state in two
steps: an isobaric expansion at P
i
followed by an isochoric (isovolumetric) process at volume V
f
. The amount of heat supplied to the system in the two-step process is approximately

Accepted Answer

In the first process : ${p_i}p_i^\gamma  = {p_f}V_f^\gamma $
$ \Rightarrow {{{p_i}} \over {{p_f}}} = {\left( {{{{V_f}} \over {{V_i}}}} ight)^\gamma } \Rightarrow 32 = {8^\gamma }$
$\gamma  = {5 \over 3}$ ... (i)
For the two step process,
$W = {p_i}({V_f} - {V_i}) = {10^5}(7 	imes {10^{ - 3}})$
$W = 7 	imes {10^2}J$
$\Delta U = {f \over 2}({p_f}{V_f} - {p_i}{V_i})$
$ = {1 \over {\gamma  - 1}}\left( {{1 \over 4} 	imes {{10}^2} - {{10}^2}} ight)$
$\Delta U =  - {3 \over 2}.{3 \over 4} 	imes {10^2} =  - {9 \over 8} 	imes {10^2}J$
$Q - W = \Delta U$
$ \Rightarrow Q = 7 	imes {10^2} - {9 \over 8} 	imes {10^2}$
$ = {{47} \over 8} 	imes {10^2}J = 588J$
![](https://imagex.cdn.examgoal.net/1kle3xr27/419ef763-622c-4815-9991-0cbed0a11df8/be5bfcf0-73b0-11eb-866f-cf1a7ff35e0c/file-1kle3xr28-320w.jpeg)

Answer

112 J

Answer

294 J

Answer

588 J

Answer

813 J

Thermodynamics & KTG

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