I would appreciate some help.
1 answer:
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Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
= 1 m³
= 10°C = 283 K
= 350 kPa
= 3 kg
= 35°C = 308 K
= 150 kPa
Now, lets apply the ideal gas equation;
=
R
=
R
/
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
= ( 3 × 0.287 × 308) / 150
= 265.188 / 150
= 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, =
/ R
= (350 × 1)/(0.287 × 283) = 350 / 81.221
= 4.309 kg
Total mass, =
+
= 4.309 + 3 = 7.309 kg
Total volume =
+
= 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
=
R
/
given that; final temperature = 20°C = 293 K
we substitute
= ( 7.309 × 0.287 × 293) / 2.77
= 614.6211119 / 2.77
= 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa