Question:medium

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

Show Hint

For an ideal gas, an isobar is a graph of Volume (V) vs. Temperature (T) at constant pressure (P). The graph is a straight line through the origin ($V = (R/P)T$), and the slope is $R/P$. Remember that the slope of this graph is inversely proportional to the pressure.

Updated On: Mar 30, 2026

$m_1>m_2>m_3$
$m_1<m_2<m_3$
$m_1>m_3>m_2$
$m_1 = m_2 = m_3$

Show Solution

The Correct Option is A

Solution and Explanation

Was this answer helpful?

0

Top Questions on States of matter

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra