A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while losing an amount q of heat to the surroundings at temperature T. If the gas constant of the gas is R, the entropy change of the gas âs during this process is:a. âS =R ln (P2/P1)b. âS = R ln (P1/P2)c. âS =R ln (P2/P1) - q/Td. âS = R ln (P1/P2) - q/Te. 0

Option B

Change in entropy of the process is

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

recall from first law,

hence we have,

since the process is isothermal,

this gives us

integrating within the limits of 1 and 2, will give us

also from ideal gas laws,

hence we have    

This makes the correct option B

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

not moving. pos accel,