Question 199: Thermodynamics & KTG

Question

A poly-atomic molecule $\left(C_V=3 R, C_P=4 Right.$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 	imes 10^{-6} \mathrm{~m}^3ight)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 	imes 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 	imes 10^{-6} \mathrm{~m}^3ight)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4ight.$ $\mathrm{Pa}, \mathrm{V}_C=8 	imes 10^{-6} \mathrm{~m}^3$ ). A to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path.
The net heat absorbed per unit mole by the system is :

Accepted Answer

As A to B is can adiabatic path,
So, $$\Delta {Q_{A 	o B}} = o$$ .... (1)
So, now we need to find the heat absorbed for the isothermal path (B to C)
We know, change in internal energy
$$\Delta u = m{c_v}\Delta T$$
$$ \Rightarrow \Delta u = o$$ (for isothermal as $$\Delta T = o$$)
Now, using Ist law of thermodynamics
$$\Delta Q = \Delta u + w$$
$$ \Rightarrow \Delta Q = o + w \Rightarrow \Delta Q = w$$
So, $$\Delta {Q_{B 	o C}} = {W_{B 	o C}}$$
$$ = nRT\ln \left( {{{{v_C}} \over {{v_B}}}} ight)$$
$$ \Rightarrow \Delta {Q_{B 	o C}} = nR(450)\ln \left( {{4 \over 3}} ight)$$
$$ \Rightarrow \Delta {Q_{B 	o C}} = 450nR\ln \left( {{4 \over 3}} ight)$$ .... (2)
So net heat absorbed,
$$Q = \Delta {Q_{A 	o B}} + \Delta {Q_{B 	o C}}$$
$$ \Rightarrow Q = 450nR\ln \left( {{4 \over 3}} ight)$$ ... (from 1 and 2)
Net heat absorbed per mole,
$$ \Rightarrow {Q \over n} = 450R(\ln 4 - \ln 3)$$.
Option 2 is correct.

Answer

$500 \mathrm{R}(\ln 3+\ln 4)$

Answer

$450 \mathrm{R}(\ln 4-\ln 3)$

Answer

$500 \mathrm{R} \ln 2$

Answer

$400 \mathrm{R} \ln 4$

Thermodynamics & KTG

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