Question

Show that the slope of the p−V diagram is greater for an adiabatic process compared to an isothermal process.

Answer 1

Explanation:

Step 1:

For an isothermal process

$P V=k \quad \ldots \text { eq } ^n(i)$

On differentiating it with respect to  V, we get

$\begin{aligned}&\frac{V d P}{d V}+P=0\\&\frac{d P}{d V}=-\frac{P}{V}\\&\frac{d P}{d V}=-\frac{k}{V 2}[Using(i)]\end{aligned}$

k is  constant.

Step 2:

method of  adiabatic process,  

$p V^{'}=K \ldots . \text { eq }^{n} \text {(ii)}$

Step 3:

On differentiating it with respect to  V, then we get

$\begin{aligned}&V^{\gamma} \frac{d P}{d V}+\gamma^{p V \gamma-1}=0\\&\frac{d P}{d V}=-\frac{\gamma^{p v^{\gamma-1}}}{v^{\gamma+1}} \quad[\text {Using}(\text {ii})], p>1 \text { and }\end{aligned}$

K = constant

γ and $\frac{dP}{dV}$ are curve slopes and heat efficiency ratios at constant pressure and volume, respectively;

P is pressure, and

V is device volume.

When you equate the two slopes, Keep in mind the γ >1, we could that see that the slope of the P-V diagram is greater than That of an adiabatic isothermic phase.

Answer 2

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Please refer the above attachment