The isobars of one mole of an ideal gas were | vedclass

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

Question

(A) For one mole of an ideal gas, the ideal gas equation is $pV = RT$, which can be rearranged as $V = (R/p)T$. Comparing this with the equation of a

Vedclass · Accepted Answer

$m_1 > m_2 > m_3$

Vedclass · Answer

(A) For one mole of an ideal gas, the ideal gas equation is $pV = RT$, which can be rearranged as $V = (R/p)T$. Comparing this with the equation of a straight line $y = mx + c$, where $y = V$ and $x = T$, the slope $m$ is given by $m = R/p$. Since $R$ is a constant, the slope $m$ is inversely proportional to the pressure $p$ $(m \propto 1/p)$. Given the condition $p_1  1/p_2 > 1/p_3$. Therefore, the relation between the slopes is $m_1 > m_2 > m_3$.

The isobars of one mole of an ideal gas were obtained at three different pressures ($p_1$, $p_2$ and $p_3$). The slopes of these isobars are $m_1$, $m_2$ and $m_3$ respectively. If $p_1 < p_2 < p_3$, then the correct relation of the slopes is

Similar Questions

Which of the following expressions at constant pressure represents Charles's law?

At $300 \ K$ temperature,the pressure of $2.5 \ g$ of $N_2$ gas is $4 \ bar$ and its volume is $2.5 \ L$. If the pressure becomes $10 \ bar$ at the same temperature,calculate its new volume. (in $L$)

Under what conditions does a gas behave ideally? Why?

If the density ratio of a gas is $1:2$ and the temperature ratio is $2:1$,what will be the ratio of their pressures?

The value of the gas constant $R$ depends on which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module