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Air in a tank is at 300 kPa, 400 K with a volume of 2 m3. A valve on the tank is opened to let...

Question:

Air in a tank is at 300 kPa, 400 K with a volume of 2 {eq}m^3 {/eq}. A valve on the tank is opened to let some air escape to the ambient to a final pressure inside of 200 kPa. Find the final temperature and mass assuming a reversible adiabatic process for the air remaining inside the tank.

Answer and Explanation:


Given data:

  • The initial temperature in the tank is: {eq}{T_1} = 400\;{\rm{K}} {/eq}
  • The initial pressure in the tank is: {eq}{P_1} = 300\;{\rm{kPa}} {/eq}
  • The volume of the tank is: {eq}V = 2\;{{\rm{m}}^3} {/eq}
  • The final pressure in the tank is: {eq}{P_2} = 200\;{\rm{kPa}} {/eq}


Write the expression for the final temperature for adiabatic process.

{eq}{T_2} = {T_1}{\left( {\dfrac{{{P_2}}}{{{P_1}}}} \right)^{\dfrac{{\gamma - 1}}{\gamma }}} {/eq}

Here, the adiabatic index is {eq}\gamma {/eq} and its value for air is {eq}1.4 {/eq}.

Substitute the values in the above equation.

{eq}\begin{align*} {T_2} &= 400\;{\rm{K}}{\left( {\dfrac{{200\;{\rm{kPa}}}}{{300\;{\rm{kPa}}}}} \right)^{\dfrac{{1.4 - 1}}{{1.4}}}}\\ &= 356.244\;{\rm{K}} \end{align*} {/eq}


Thus the final temperature is {eq}356.244\;{\rm{K}} {/eq}.


Write the expression for the final mass of the air in the tank.

{eq}m = \dfrac{{{P_2}V}}{{R{T_2}}} {/eq}

Here, the characteristic gas constant is {eq}R {/eq} and its value for air is {eq}0.287\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}} {/eq}.

Substitute the values in the above equation.

{eq}\begin{align*} m &= \dfrac{{200 \times 2}}{{0.287 \times 356.244}}\\ &= 3.9122\;{\rm{kg}} \end{align*} {/eq}


Thus the final mass of the air in the tank is {eq}3.9122\;{\rm{kg}} {/eq}.


Learn more about this topic:

Thermodynamic Processes: Isobaric, Isochoric, Isothermal & Adiabatic

from TExES Physics/Mathematics 7-12 (243): Practice & Study Guide

Chapter 55 / Lesson 3
132K

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