a) Sketches of all possible pv-diagrams for the cycle are attached below
b) The work W
for the process Ca is : 2462.8 J
<u>Given data :</u>
Amount of heat flowing out = 800 J
Ta = 200 K
Tb = 300 K
R = 800
<u>B) Determine the </u><u>work W </u><u>for the process</u><u> Ca</u><u> </u>
Wₐs = -pdv
= - [ pVb - pVa ] ---- ( 1 )
note : pVb = nRTb , pVa = nRTa
Equation ( 1 ) becomes
= -nR [ Tb - Ta ]
= - 2(8.314 ) [ 300 - 200 ]
= - 1662.87
given that W
= 0 which is isochonic
dv = 0 ( cyclic process ) = d∅ - dw
∴ 0 = 800 - ( Wₐs + W )
Therefore : W = 800 + 1662.8 = 2462.8 J
Hence we can conclude that the work W for the process Ca = 2462.8 J
Learn more about Pv diagrams : brainly.com/question/25401637
Answer:
What is quantum physics? Put simply, it's the physics that explains how everything works: the best description we have of the nature of the particles that make up matter and the forces with which they interact. Quantum physics underlies how atoms work, and so why chemistry and biology work as they do.
Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
