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Rashid [163]
3 years ago
15

If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

hat happened to the pressure if the temperature remains the same. Include which law supports the observation
Physics
1 answer:
Ann [662]3 years ago
3 0

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: P*V=n*R*T

where P is the gas' pressure, V its volume, n the number of moles of gas,  R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450

So the pressure must have gone up to 450 kilopascals.

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A gas bottle contains 0.250 mol of gas at 730 mm hg pressure. if the final pressure is 1.15 atm, how many moles of gas were adde
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Answer: 0.049 mol



Explanation:



1) Data:


n₁ = 0.250 mol

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p₂ = 1.15 atm

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2) Assumptions:


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ii) V and T constants.


3) Solution:


i) Since the temperature and the volume must be assumed constant, you can simplify the ideal gas equation into:


pV = nRT ⇒ p/n = RT/V ⇒ p/n = constant.


ii) Then p₁ / n₁ = p₂ / n₂


⇒ n₂ = p₂ n₁ / p₁


iii) n₂ = 1.15atm × 760 mmHg/atm × 0.250 mol / 730mmHg = 0.299 mol


iv) n₂ - n₁ = 0.299 mol - 0.250 mol = 0.049 mol

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