Correct Answer - A
(ab process) From the given graph, we can see that
`(p prop T)`
`rArr (V = "constant") (isochoric)
or `(rho = "constant") (as `rho prop 1/v)`
`(p)` and `(T)` both are increasing.
Therefore, U is also increasing ( as `u prop T)`
Now,
(i) (p - V) graph is straight line parallel to (p - axis) as (V) is constant.
(ii) (V - T) graph is a straight line parallel to (T - axis) as (V) is constant.
(iii) (p - T) graph is a straight line parallel to (T - axis) as (rho) is constant.
(iv) (U - T) graph is a straight line passing through origin as `( U prop T)`.
(bc process) From the given graph, we can see that
(T = constant) (isothermal)
rArr (U = constant)
` rArr (pV = "constant")`
or `p prop 1/V`
(p) is increasing. Therefore, (V) will decrease.
Hence, `(rho)` will increase.
now,
(i) (p - V) graph is a rectangular hyperbola ( as P prop 1/V).
(ii) `(V - T)` graph is a straight line parallel to (`V` - axis), as (`T =` constant)
(iii) `(rho - T)` graph is a striaght line parallel to ( rho - axis), as (T = constant)
(iv) ( U - T) graph is a dot, as (U) and (T) both are constants.
(cd process) From the given graph, we can see that
( p = constant)
rArr ( V prop T) (isobaric)
Temperature is decreasing. So, volume will also decrease. But density will increase.
Now,
(i) (p - V) graph is a straight line parallel to (V - axis) because (p) is constant.
(ii) (V - T) graph is a straight line passing through origin, as ( V prop T).
Now,
(i) (p - V) graph is a straight line parallel to (V - axis) because (p) is constant.
(ii) (V - T) graph is a straight line passing through origin, as ( V prop T).
(iii) ` rho = (pM)/(RT) rArr rho prop 1/T` as (p), (M) and (R) al are constants. Hence, `(rho - T)` graph is rectangular hyperbola.
(iv) (U - T) graph is a straight line passing through origin as `( U prop T)`.
(da process) This process is just inverse of (bc) process. So, this process will complete the cycle following the steps discussed in process (bc).
The four graphs are as shown below.
