Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressures P1 and P2. Then,

Question

Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressures P1 and P2. Then, (A) P1 > P2 (B) P1 < P2 (C) P1=P2 (D) The information is insufficient

Tardigrade Admin · Accepted Answer

Since the graph is a straight line, so, V=mT where m is the slope. =(n R T) / P [From equation of state] or m propto(1 / P) or, m2 > m1 ∴ P2 < P1

Q. Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressures $P_{1}$ and $P_{2}$ . Then,

$P_{1} > P_{2}$

$P_{1} < P_{2}$

$P_{1} = P_{2}$

The information is insufficient

Since the graph is a straight line,
so, $V = m T$ where $m$ is the slope.
$= (n RT) / P$ [From equation of state]
or $m \propto (1/ P)$ or, $m_{2} > m_{1}$
$∴ P_{2} < P_{1}$

Q. Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressures P1​ and P2​. Then,

P1​>P2​

P1​<P2​

P1​=P2​