Question 829: Thermodynamics & KTG

Question

In a thermodynamic process, there is no exchange of heat between the system and surroundings. Then the thermodynamic process is

Accepted Answer

According to Wien's law, $$\lambda_{\mathrm{m}} \mathrm{T}=$$ constant (b)
$$	herefore \mathrm{T}=\frac{\mathrm{b}}{\lambda_{\mathrm{m}}}\quad 	ext{..... (1)}$$
and from Stefan's law, $$\mathrm{Q}=\sigma \mathrm{AT}^4\quad 	ext{.... (2)}$$
For the disc, area $$(\mathrm{A})=\pi \mathrm{r}^2$$
$$	herefore$$ From (1) and (2),
$$\mathrm{Q}=\sigma \cdot \pi \mathrm{r}^2 \cdot \frac{\mathrm{b}^4}{\lambda_{\mathrm{m}}^4}=\frac{\mathrm{Kr}^2}{(\lambda \mathrm{m})^4}$$
where $$\mathrm{K}=\pi \sigma \mathrm{b}^4$$ is a constant
$$\mathrm{Q}$$ is the quantity of heat radiated per second or power.
Hence $$\mathrm{P}_{\mathrm{x}}, \mathrm{P}_{\mathrm{y}}$$ and $$\mathrm{P}_{\mathrm{z}}$$ are the powers of $$\mathrm{x}, \mathrm{y}, \mathrm{z}$$.
For $$\mathrm{x}$$ we have $$\mathrm{r}_1=2 \mathrm{~m}$$ and $$\lambda_1=300 \mathrm{~nm}$$
For $$y$$, we have $$r_2=3 \mathrm{~m}$$ and $$\lambda_2=400 \mathrm{~nm}$$
and For $$\mathrm{z}$$, we have $$\mathrm{r}_3=6 \mathrm{~m}$$ and $$\lambda_3=500 \mathrm{~nm}$$
$$\begin{aligned}
	herefore \mathrm{P}_{\mathrm{x}} & \propto \frac{\mathrm{r}_1^2}{\lambda_1^4} 	ext { or } \frac{2^2}{\left(3 	imes 10^{-7}ight)^4} 	ext { or } \frac{4 	imes 10^{+28}}{81} \
\mathrm{P}_{\mathrm{y}} & \propto \frac{\mathrm{r}_2^2}{\lambda_2^4} 	ext { or } \frac{9 	imes 10^{+28}}{256} \
\mathrm{P}_{\mathrm{z}} & \propto \frac{\mathrm{r}_3^2}{\lambda_3^4} 	ext { or } \frac{36 	imes 10^{28}}{625}
\end{aligned}$$
But $$\frac{4}{81}=0.049, \frac{9}{250}=0.035$$ and $$\frac{36}{625}=0.0576$$
$$	herefore \mathbf{P}_{\mathrm{z}}$$ is maximum.

Answer

isothermal

Answer

isobaric

Answer

isochoric

Answer

adiabatic

Thermodynamics & KTG

In a thermodynamic process, there is no exchange of heat between the system and surroundings. Then the thermodynamic process is

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