An engine takes in 5 moles of air at
Question:

An engine takes in 5 moles of air at $20^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$, and compresses it adiabaticaly to $1 / 10^{\mathrm{th}}$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be $\mathrm{X} \mathrm{kJ}$. The value of $\mathrm{X}$

to the nearest integer is

 

 

Solution:

Diatomic :

$\mathrm{f}=5$

$\gamma=7 / 5$

$\mathrm{~T}_{\mathrm{i}}=\mathrm{T}=273+20=293 \mathrm{~K}$

$\mathrm{~V}_{\mathrm{i}}=\mathrm{V}$

$\mathrm{V}_{\mathrm{f}}=\mathrm{V} / 10$

Adiabatic

TV $\gamma-1=$ constant

$\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$

$\mathrm{T} \cdot \mathrm{V}^{7 / 5-1}=\mathrm{T}_{2}\left(\frac{\mathrm{V}}{10}\right)^{7 / 5-1}$

$\Rightarrow \mathrm{T}_{2}=\mathrm{T} \cdot 10^{2 / 5}$

$\Delta U=\frac{n f R\left(T_{2}-T_{1}\right)}{2}=\frac{5 \times 5 \times \frac{25}{3} \times\left(T .10^{2 / 5}-T\right)}{2}$

$=\frac{25 \times 25 \times}{6} \mathrm{~T}\left(10^{2 / 5}-1\right)$

$=\frac{625 \times 293 \times\left(10^{2 / 5}-1\right)}{6}$

$=4.033 \times 10^{3} \approx 4 \mathrm{~kJ}$

 

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