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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases

In a thermodynamic process, an ideal monoatomic gas follows the law $TP^{\large\frac{3}{5}} = $ constant. If temperature of 2 moles of gas is raised from T to 4 T, then what is the heat given to the gas?

$\begin{array}{1 1} 12.5\;R \\ 25\;R \\ 50 R \\ 100 R \end{array}$

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1 Answer

For an ideal gas, $PV = nRT$
Here, its given that $TP^{\large\frac{3}{5}} = $ constant $\rightarrow P^{\large\frac{8}{5}}$$V = k = $ constant.
$\Rightarrow P = KV^{\large\frac{-5}{8}}$
Differentiating the equation: $V (KV^{\large\frac{-5}{8}})$$ = nRT$, we get:
$\large\frac{3K}{8}$$V^{\large\frac{-5}{8}}$$dV = nRDT$
$\Rightarrow PdV = \large\frac{8}{3}$$nrDT = \Delta w$
Now, $\Delta Q = \Delta u + \Delta w \rightarrow \Delta Q = n \large\frac{3R}{2}$$\Delta T + \large\frac{8nR}{3}$$\Delta T$
$\Rightarrow \Delta Q = nR\Delta T \large \frac{25}{6}$$ = 25 R$
answered Jan 23, 2014 by balaji.thirumalai
 

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