The graph of four gases with the same number | vedclass

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

Question

(A) From the ideal gas equation,$PV = \mu RT$,we can write:$P = \left( \frac{\mu R}{V} \right) T$Comparing this with the equation of a straight line $

Vedclass · Accepted Answer

$V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$

Vedclass · Answer

(A) From the ideal gas equation,$PV = \mu RT$,we can write:$P = \left( \frac{\mu R}{V} ight) T$Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m$ is given by:$m = 	an 	heta = \frac{\mu R}{V}$Since the number of moles $\mu$ and the gas constant $R$ are constant,we have:$V \propto \frac{1}{	an 	heta}$As the angle $	heta$ increases,$	an 	heta$ increases,and therefore the volume $V$ decreases.From the graph,the slope of line $1$ is equal to the slope of line $2$ (as they are parallel),so $V_1 = V_2$.Similarly,the slope of line $3$ is equal to the slope of line $4$,so $V_3 = V_4$.Since the slope of line $3$ (and $4$) is greater than the slope of line $1$ (and $2$),we have $	an 	heta_3 > 	an 	heta_1$.Therefore,$V_3  V_3$.Thus,the correct relation is $V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$.

The graph of four gases with the same number of moles is given. Then:

Similar Questions

The pressure $(P)$ and temperature $(T)$ relationship of an ideal gas obeys the equation $PT^2 = \text{constant}$. The volume expansion coefficient of the gas will be:

Under constant temperature,the graph between $P$ and $1/V$ is:

What is the value of the volume of an ideal gas at absolute zero temperature?

The $P-T$ graph shows a gas being heated. During this process from point $1$ to $2$,the volume of the gas:

At $100 \, K$ and $0.1 \, \text{atm}$ pressure, the volume of helium gas is $10 \, \text{litres}$. If volume and pressure are doubled, its temperature will change to ....... $K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module