Chemistry, 14.11.2019 02:31 jimmymurray29

Arigid, well-insulated tank is filled initially with 5.0 kg of air at a pressure of 5 bars and a temperature of 500k. a leak develops, and air slowly escapes until the pressure of the air remaining in the tank is 1 bar. employing the ideal gas model, determine the amount of mass remaining in the tank and its temperature.

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Answer from: cddominguez63

Explanation:

In an isotropic process, the equation is as follows.

$\frac{T_{2}}{T_{1}} = [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}$

$T_{2} = T_{1} \times [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}$

As the given data is as follows.

$T_{1}$ = 500 K, m = 5 kg

$P_{1}$ = 5 bar, $P_{2}$ = 1 bar

Now, putting the given values into the above formula as follows.

$T_{2} = T_{1} \times [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}$

$T_{2} = 500 K \times [\frac{5 bar}{1 bar}]^{\frac{1.4 - 1}{1.4}}$

= 315.7 K

As according to the ideal gas equation, PV = mRT

So, calculate the volume as follows.

V = $\frac{mRT}{P}$

= $\frac{5 kg \times 287 J/kg \times 500 K}{5 \times 10^{5}}$ (as 1 bar = 10^{5} Pa[/tex])

= $1.435 m^{3}$

Now, we will calculate the value of $m_{2}$ as follows.

$P_{2}V_{2} = m_{2}RT_{2}$

$m_{2} = \frac{P_{2}V_{2}}{RT_{2}}$

= $\frac{1 \times 10^{5} \times 1.435 m^{3}}{315.7 K \times 287 J/kg}$

= 1.58 kg

Thus, we can conclude that the amount of mass remaining in the tank is 1.58 kg and its temperature is 315.7 K.

Answer from: Quest

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Answer from: Quest

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