Question

A 1-m3 tank containing air at 25 °C and 500 kPa is connected through a valve to another tank containing 5 kg of air at 35 °C and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20 °C. Determine the volume of the second tank and the final equilibrium pressure of air.

Answer 1

Answer:

Volume of the second tank is 3.21m³

Final eguilibrium presssure is 284.13kpa

Explanation:

Volume of the tank A, $V_A=1m^3$

Temperature of air in tank A,$T_A = 25^0C = 298K$

Pressure of air in tank A, P_A = 500kPa

Mass of air in tank B, m_b = 5kg

Temperature of air in tank B,$T_B = 35^0C = 308k$

Pressure of air in tank B, $P_B = 200kPa$

Surrounding temperature,$T_s_u_r_r = 20^0c = 293K$

Assuming, at given conditions air behaves as an ideal gas.

For air, gas constant R = 0.287 kJ/kmol K

From ideal gas equation, mass of air in tank A is determined by

$m_A = \frac{P_AV_A}{RT_A} \\\\m_A = \frac{500\times1}{0.287\times298} \\\\m_A = 5.846kg$

Volume of the tank B can be determined from

$V_B=\frac{m_BRT_B}{P_B} \\\\V_B = \frac{5\times0,287\times308}{200} \\\\V_B = 2.21m^3$

so,when the valve is opened

Total volume,

$V = V_A+V_B\\V= 1+2.21\\V=3.21m^3$

Volume of the second tank is 3.21m³

Total mass of air,

$m= m_A+m_B\\m = 5+5.846\\m= 10.846kg$

The final equilibrium pressure (P) can be obtained from the ideal gas equation applied to total volume

$P = \frac{mRT_s_u_r_r}{V} \\\\P=\frac{10.846\times0.287\times293}{3.21} \\\\P= 284.13kpa$

Final eguilibrium presssure is 284.13kpa