Pressure versus temperature graphs of an idea | vedclass

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(D) For an ideal gas, the equation of state is $PV = nRT = (m/M)RT$, where $m$ is the mass and $M$ is the molar mass. Since density $\rho = m/V$, we c

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) For an ideal gas, the equation of state is $PV = nRT = (m/M)RT$, where $m$ is the mass and $M$ is the molar mass. Since density $ho = m/V$, we can write $P = (ho/M)RT$, which implies $ho = (PM)/(RT)$.In graph $(i)$, $T$ is constant and $P$ is increasing. Since $ho \propto P/T$, if $P$ increases at constant $T$, density $ho$ increases. Thus, statement $(A)$ is correct.In graph $(ii)$, $P$ is constant and $T$ is increasing. Since $ho \propto P/T$, if $T$ increases at constant $P$, density $ho$ decreases. Thus, statement $(B)$ is correct.In graph $(iii)$, the graph is a straight line passing through the origin, so $P/T = 	ext{constant}$. Since $ho = (PM)/(RT)$, if $P/T$ is constant, then density $ho$ is constant. Thus, statement $(C)$ is correct.Since all statements $(A), (B),$ and $(C)$ are correct, the wrong statement is none of the above.

Pressure versus temperature graphs of an ideal gas are as shown in the figure. Choose the wrong statement.

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