The state of an ideal gas was changed isobari | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of state for an ideal gas is $pV = nRT$,which can be rearranged as $V = (nR/p) \cdot T$.This represents the equation of a straight li

Vedclass · Accepted Answer

$p_1 > p_2 > p_3$

Vedclass · Answer

(B) The equation of state for an ideal gas is $pV = nRT$,which can be rearranged as $V = (nR/p) \cdot T$.This represents the equation of a straight line passing through the origin with a slope $m = nR/p$.From the relationship,we can see that the slope is inversely proportional to the pressure,i.e.,$	ext{slope} \propto 1/p$.By observing the given $V-T$ graph,the slope of the line for $p_3$ is the greatest,followed by $p_2$,and the slope for $p_1$ is the smallest.Since $	ext{slope} \propto 1/p$,a larger slope corresponds to a smaller pressure.Therefore,the relationship between the pressures is $p_1 > p_2 > p_3$.

$A$. Process $A$	$P=5, V=7$
$B$. Process $B$	$P=4, V=6$
$C$. Process $C$	$P=12, V=1$
$D$. Process $D$	$P=6, V=3$

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?

Similar Questions

The average force applied on the wall of a closed container depends as $T^{x}$ where $T$ is the temperature of an ideal gas. The value of $x$ is

In Boyle's law, what remains constant?

By using the ideal gas equation $PV = \mu RT$,write the value,unit,and dimension of $R$.

$S.I.$ unit of universal gas constant is

Vedclass Test Series

Exam Paper Generator

Online Exam Module