Pressure versus temperature graph of an ideal gas is shown in figure. Density of the gas at point A is `rho_(0)`. Density at B will be

Options

A. `2 rho_(0) `
B. `(2)/(3) rho_(0)`
C. `(3)/(2) rho_(0)`
D. `3 rho_(0)`

Answer

B

Explanation

Given, density of gas at point `A, rho_(A) = rho_(0)` General equation for an ideal gas
`pV= nRT`
`implies pV = (m)/(M) . RT [ m rarr ` mass M `rarr` molecular mass ]
`implies p = (m)/(V). (R )/(M ) . T`
`implies p =rho(( R)/(M))T " " [ "where density " rho = (m)/(V) ] cdots `
Applying Eq. (i) at point A and B, respectively
`rho =rho_(A) ((R)/(M)) T_(0)`
`2p = rho_(B)((R)/(M)) 3 T_(0)`
From Eqs . (ii) and (iii) we get
`(1)/(2) = (rho_(A))/(rho_(B)). (1)/(3)`
`implies rho_(B) = (2)/(B) rho_(A) = (2)/(3) rho_(0)`