The state of an ideal gas was changed isobarically. The graph depicts three such isobaric lines. Which of the following is true about the pressures of the gas?

Question

(A) p1 = p2 = p3 (B) p1> p2> p3 (C) p1 < p2 < p3 (D) p1/p2 = p3 /p1

Tardigrade Admin · Accepted Answer

Equation of state for ideal gas is pV=nRT⇒ V=(nR/p).T This is equation of line with slope =(nR/p) ⇒ Slope propto (1/p) So, from graph, p3 < p2 < p1

Q. The state of an ideal gas was changed isobarically. The graph depicts three such isobaric lines. Which of the following is true about the pressures of the gas?

$p_{1} = p_{2} = p_{3}$

$p_{1} > p_{2} > p_{3}$

$p_{1} < p_{2} < p_{3}$

$p_{1} / p_{2} = p_{3} / p_{1}$

Equation of state for ideal gas is
$p V = n RT \Rightarrow V = \frac{n R}{p} . T$
This is equation of line with slope $= \frac{n R}{p}$
$\Rightarrow$ Slope $\propto \frac{1}{p}$
So, from graph, $p_{3} < p_{2} < p_{1}$

Q. The state of an ideal gas was changed isobarically. The graph depicts three such isobaric lines. Which of the following is true about the pressures of the gas?

p1​=p2​=p3​

p1​>p2​>p3​

p1​<p2​<p3​

p1​/p2​=p3​/p1​

Equation of state for ideal gas is pV=nRT⇒V=pnR​.T This is equation of line with slope =pnR​ ⇒ Slope ∝p1​ So, from graph, p3​<p2​<p1​

$p_{1} = p_{2} = p_{3}$

$p_{1} > p_{2} > p_{3}$

$p_{1} < p_{2} < p_{3}$

$p_{1} / p_{2} = p_{3} / p_{1}$

Equation of state for ideal gas is
$p V = n RT \Rightarrow V = \frac{n R}{p} . T$
This is equation of line with slope $= \frac{n R}{p}$
$\Rightarrow$ Slope $\propto \frac{1}{p}$
So, from graph, $p_{3} < p_{2} < p_{1}$