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A balloon contains 500 m3 of helium at 27°C and 1 atmosphere pressure. The volume of the helium at —3° C temperature and 0.5 atmosphere pressure will be
- a)1000 m3
- b)900 m3
- c)700 m3
- d)500 m3
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A balloon contains 500 m3of helium at 27°C and 1 atmosphere pressu...
U can use combined gas formula... p1 v1 / t1 = p2 v2 /t2
where p1 = 1 , t1= 300k, v1= 500
p2 = 0.5. , t2 = 270 k , v2 =?
n u wil get the required ans.
where p1 = 1 , t1= 300k, v1= 500
p2 = 0.5. , t2 = 270 k , v2 =?
n u wil get the required ans.
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Community Answer
A balloon contains 500 m3of helium at 27°C and 1 atmosphere pressu...
°C and 1 atm pressure. If the balloon rises to an altitude where the temperature drops to 10°C and the pressure is 0.8 atm, what is the new volume of the balloon?
Assuming the balloon and the gas inside it behave ideally, we can use the ideal gas law to solve for the new volume:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature.
Since the number of moles of helium does not change, we can write:
PV/T = constant
At the initial conditions, we have:
P1 = 1 atm
V1 = 500 m3
T1 = 27°C = 300 K (adding 273 to convert from Celsius to Kelvin)
R = 8.31 J/(mol·K) (the ideal gas constant)
So we can solve for the initial number of moles:
n = P1 V1 / (R T1) = (1 atm) (500 m3) / (8.31 J/(mol·K) × 300 K) ≈ 20,000 moles
Now, at the new conditions, we have:
P2 = 0.8 atm
T2 = 10°C = 283 K
n = 20,000 moles
We can solve for the new volume:
V2 = n R T2 / P2 = (20,000 moles) (8.31 J/(mol·K) × 283 K) / (0.8 atm) ≈ 707 m3
Therefore, the new volume of the balloon is approximately 707 m3.
Assuming the balloon and the gas inside it behave ideally, we can use the ideal gas law to solve for the new volume:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature.
Since the number of moles of helium does not change, we can write:
PV/T = constant
At the initial conditions, we have:
P1 = 1 atm
V1 = 500 m3
T1 = 27°C = 300 K (adding 273 to convert from Celsius to Kelvin)
R = 8.31 J/(mol·K) (the ideal gas constant)
So we can solve for the initial number of moles:
n = P1 V1 / (R T1) = (1 atm) (500 m3) / (8.31 J/(mol·K) × 300 K) ≈ 20,000 moles
Now, at the new conditions, we have:
P2 = 0.8 atm
T2 = 10°C = 283 K
n = 20,000 moles
We can solve for the new volume:
V2 = n R T2 / P2 = (20,000 moles) (8.31 J/(mol·K) × 283 K) / (0.8 atm) ≈ 707 m3
Therefore, the new volume of the balloon is approximately 707 m3.
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A balloon contains 500 m3of helium at 27°C and 1 atmosphere pressure. The volume of the helium at —3° C temperature and 0.5 atmosphere pressure will bea)1000 m3b)900 m3c)700 m3d)500 m3Correct answer is option 'B'. Can you explain this answer?
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