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Physics Class 12th Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Six

The change in the magnitude of the volume of an ideal gas when a small additional pressure $Δ P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $Δ T$ at constant pressure. The initial temperature and pressure of the gas were $300 K$ and 2 atm. respectively. If $| Δ T | = C | Δ P |$ then value of $C$ in (K/atm) is ....

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alok kumar singh | Contributor-Level 10

3 months ago

Sol.

$\begin{array}{l} A H \to \frac{H}{2} & H = \frac{\sqrt[]{3} a}{2} \\ A G \to \frac{H}{\sqrt[]{3}} & A l = \frac{H}{2 \sqrt[]{3}} \\ M_{A B C} = M & M_{A D E} = \frac{M}{4} \end{array}$

$\Rightarrow I_{G} = \frac{M a^{2}}{12} - [\frac{M}{4} \frac{{[\frac{a}{2}]}^{2}}{12} + \frac{M}{4} {[\frac{a}{2 \sqrt[]{3}}]}^{2}] = (\frac{11}{16}) \frac{{M a}^{2}}{12}$

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Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Process-AB $\Rightarrow$ Isobaric,

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from newtan’s law of cooling :

$\frac{d T}{d t} = - k (T - T_{0})$

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

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Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = 50$

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$r = \frac{m v}{q B}$

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