A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume increases to 30.0 ml, at constant temperature is 0.25atm.
On constant temperature, the pressure and volume relation become constant before and after the change in quantitities have occurred.
According to Boyle's Law,
P₁V₁ = P₂V₂
where, P₁ is pressure exerted by the gas initially
V₁ is the volume of gas initially
P₂ is pressure exerted by the gas finally
V₂ is the volume of gas finally
Given,
P₁ = 1.5 atm
V₁ = 5 ml
V₂ = 30 ml
P₂ =?
On substituting the given values in the above equation:
P₁V₁ = P₂V₂
1.5 atm × 5 ml = P₂ × 30 ml
P₂ = 0.25 atm
Hence, pressure exerted by the gas is 0.25atm.
Learn more about Boyle's Law here, brainly.com/question/1437490
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Weight. Because there is less gravity on the moon.
Answer:
the amount or number of a material or immaterial thing not usually estimated by spatial measurement
Answer:
Amplitude will be equal to 0.091 m
Explanation:
Given mass of the slits = 41 gram = 0.041 kg
Frequency f = 1.65 Hz
So angular frequency 
Angular frequency is equal to 

Squaring both side

k = 4.40 N/m
For vertical osculation


A = 0.091 m
So amplitude will be equal to 0.0391 m
<h3><u>Given</u><u>:</u><u>-</u></h3>
Force,F = 100 N
Acceleration,a = 5 m/s²
<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>
Calculate the mass of the box .
<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

<h3><u>Solution:-</u><u> </u></h3>

★ Substituting the values in the above formula,we get:


