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A balloon contains 1500 m3 of helium at 27oC and 4 atmospheric pressure. The volume of helium at – 3oC temperature and 2 atmospheric pressure will be,
  • a)
    1500 m3
  • b)
    1700 m
  • c)
    1900 m3
  • d)
    2700 m3
Correct answer is option 'D'. Can you explain this answer?
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A balloon contains 1500 m3 of helium at 27oC and 4 atmospheric pressu...
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A balloon contains 1500 m3 of helium at 27oC and 4 atmospheric pressu...
Given:
Initial conditions:
Volume of helium (V1) = 1500 m³
Temperature (T1) = 27°C = 27 + 273 = 300 K
Pressure (P1) = 4 atm

Final conditions:
Temperature (T2) = -3°C = -3 + 273 = 270 K
Pressure (P2) = 2 atm

We need to find the final volume of helium (V2).

To solve this problem, we can use the combined gas law equation:

(P1 * V1) / T1 = (P2 * V2) / T2

where
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure
V2 = Final volume (to be determined)
T2 = Final temperature

Substituting the given values into the equation, we have:

(4 atm * 1500 m³) / 300 K = (2 atm * V2) / 270 K

We can simplify the equation by canceling out the units:

(2 * 1500) / 3 = V2 / 2

Solving for V2:

V2 = (2 * 1500 * 2) / 3
= 6000 / 3
= 2000 m³

Therefore, the final volume of helium at -3°C and 2 atm pressure is 2000 m³.
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A balloon contains 1500 m3 of helium at 27oC and 4 atmospheric pressure. The volume of helium at – 3oC temperature and 2 atmospheric pressure will be,a)1500 m3b)1700 mc)1900 m3d)2700 m3Correct answer is option 'D'. Can you explain this answer?
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