(C) From the ideal gas equation,$PV = nRT$,we have $T = (P/nR)V$. For a fixed amount of gas ($n$ is constant),the slope of the $T-V$ graph is given by

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(C) From the ideal gas equation,$PV = nRT$,we have $T = (P/nR)V$. For a fixed amount of gas ($n$ is constant),the slope of the $T-V$ graph is given by $m = P/nR$. Since $n$ and $R$ are constants,the slope $m$ is directly proportional to the pressure $P$. Therefore,a steeper slope in the $T-V$ graph corresponds to a higher pressure. Looking at the graph,the slope of the curve for $P_1$ is the steepest,followed by $P_2$,and then $P_3$. Thus,the correct relation is $P_1 > P_2 > P_3$.

Class 11 Physics Kinetic Theory of Gases Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

Medium

The following graph represents the $T-V$ curves of an ideal gas (where $T$ is the temperature and $V$ is the volume) at three pressures $P_1, P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. Then the correct relation is:

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A
$P_1 > P_3 > P_2$
B
$P_2 > P_1 > P_3$
C
$P_1 > P_2 > P_3$
D
$P_3 > P_2 > P_1$

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Class 11

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Chapter

Kinetic Theory of Gases

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Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

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Two vessels $A$ and $B$ are of the same size and are at the same temperature. Vessel $A$ contains $1 \ g$ of hydrogen and vessel $B$ contains $1 \ g$ of oxygen. If $P_{A}$ and $P_{B}$ are the pressures of the gases in $A$ and $B$ respectively,then the ratio $\frac{P_{A}}{P_{B}}$ is:

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Air is filled at $60^oC$ in a vessel with an open mouth. The vessel is heated to a temperature $T$ so that $1/4^{th}$ of the air escapes. Assuming the volume of the vessel remains constant,the value of $T$ is ....... $^oC$.

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The following graph represents the $T-V$ curves of an ideal gas (where $T$ is the temperature and $V$ is the volume) at three pressures $P_1, P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. Then the correct relation is:

Similar Questions

In Boyle's law, what remains constant?

$A$ container holds $1$ mole of $O_2$ (molar mass $32$) at temperature $T$. The pressure is $P$. Another identical container holds $1$ mole of $He$ (molar mass $4$) at temperature $2T$. Find the pressure in the second container.

Two vessels $A$ and $B$ are of the same size and are at the same temperature. Vessel $A$ contains $1 \ g$ of hydrogen and vessel $B$ contains $1 \ g$ of oxygen. If $P_{A}$ and $P_{B}$ are the pressures of the gases in $A$ and $B$ respectively,then the ratio $\frac{P_{A}}{P_{B}}$ is:

Air is filled at $60^oC$ in a vessel with an open mouth. The vessel is heated to a temperature $T$ so that $1/4^{th}$ of the air escapes. Assuming the volume of the vessel remains constant,the value of $T$ is ....... $^oC$.

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