Density `(rho)` versus internal energy (U) graph of a gas is as shown in figure. Choose the correct options.

Here, W is work done by gas and Q is heat given to the gas.

Options

A. (a) `Q_(bc)=0`
B. (b) `W_(bc)=0`
C. (c) `W_(ca)lt0`
D. (d) `Q_(ab)gt0`

Answer

C, D

Explanation

ab `rho=` constant
`:.` `V=`constant
`:.` W=0`
`Q=Delta U`
`DeltaU` is positive, as U is increasing.
Hence, Q is also positive.
bc `rho prop Uimplies :. 1/V prop T`
`rho` is decreasing, so V is increasing. Hence, work done is positive.
Further, `1/V prop T(TproppV)`
`:.` `pV^(2)=` constant
In the process, `pV^x=` constant,
Molar heat capacity is given by
`C=C_V+(R)/(1-x)`
Here, `x=2`
`:.` `C=C_V-R`
For any of the gas, `C_V!=R`.
`:.` `C!=0`
`:.` `Q=nCDeltaT!=0` as `DeltaU!=0` and `DeltaT!=0`
ca
`rho` is increasing. Hence, V is decreasing. So, work done is negative.