Answer:
(A) Work done will be 87.992 KJ
(B) Work done will be 167.4 KJ
Explanation:
We have given mass of methane m = 4.5 gram = 0.0045 kg
Volume occupies 
And volume is increased by 
so 
Temperature T = 310 K
Pressure is given as 200 Torr = 26664.5 Pa
(a) At constant pressure work done is given by
(b) At reversible process work done is given by 
We have given mass = 4.5 gram
Molar mass of methane = 16
So number of moles 
So work done 
Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
We Know, K.E. = 1/2 × m × v²
From the expression, we can conclude that Kinetic energy is directly proportional to mass. So, as mass will increase, Kinetic energy will also increase.
In short, Your Correct answer would be Option B
Hope this helps!
The answer is distressing
v^2 = v0^2 +2ad
v^2 = 22^2 + 2*3.78*45 = 824.2
v= √824.2 = 28.7 m/s
