Question 1: Thermodynamics & KTG

Question

An ideal gas undergoes a four step cycle as shown in the P-V diagram below. During this cycle, heat is absorbed by the gas in
![](https://imagex.cdn.examgoal.net/1kuckfqvb/f0e7dc7d-f3fd-4969-8399-3351b44393e9/a5f42d70-2505-11ec-b291-1df06cb45b25/file-1kuckfqvc.png)

Accepted Answer

Process 1
p = constant, Volume increases and temperature also increases
Q = W + $\Delta$U
$	herefore$ W = positive, $\Delta$U = positive
$\Rightarrow$ Heat is positive and supplied to the gas.
Process 2
V = constant, Pressure decreases
T $\propto$ pV [as V = constant]
$\Rightarrow$ Temperature decreases
W = 0
$\Delta$T is negative and $\Delta$U = ${f \over 2}$nR$\Delta$T
$	herefore$ $\Delta$U is also negative
Q = $\Delta$U + W
$	herefore$ Heat is negative and rejected by gas.
Process 3
p = constant, Volume decreases
$\Rightarrow$ Temperature also decreases
W = p$\Delta$V = negative
$\Delta$U = ${f \over 2}$nR$\Delta$T = negative
$	herefore$ Heat is negative and rejected by gas.
Process 4
V = constant, Pressure increases
W = 0 (as V = constant)
pV = nRT $\Rightarrow$ Temperature increase
$\Rightarrow$ $\Delta$U = ${f \over 2}$nR$\Delta$T is positive
$\Delta$Q = $\Delta$U + W = positive

Answer

steps 1 and 2

Answer

steps 1 and 3

Answer

steps 1 and 4

Answer

steps 2 and 4

Thermodynamics & KTG

An ideal gas undergoes a four step cycle as shown in the P-V diagram below. During this cycle, heat is absorbed by the gas in

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