The adjoining figure shows the graph of press | vedclass

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

Question

(C) For an ideal gas,the equation of state is $PV = nRT$. To compare the temperatures $T_1$ and $T_2$,we can draw a line of constant pressure $(P =

Vedclass · Accepted Answer

$T_1 < T_2$

Vedclass · Answer

(C) For an ideal gas,the equation of state is $PV = nRT$. To compare the temperatures $T_1$ and $T_2$,we can draw a line of constant pressure $(P = 	ext{constant})$ on the $P-V$ graph. At a constant pressure,the ideal gas equation gives $V \propto T$. From the graph,for a fixed pressure,the volume corresponding to the curve $T_2$ (let it be $V_2$) is greater than the volume corresponding to the curve $T_1$ (let it be $V_1$),i.e.,$V_2 > V_1$. Since $V \propto T$ at constant pressure,$V_2 > V_1$ implies $T_2 > T_1$ or $T_1 < T_2$.

The adjoining figure shows the graph of pressure $(P)$ and volume $(V)$ of a gas at two temperatures $T_1$ and $T_2$. Which of the following inferences is correct?

Similar Questions

$A$ container holds $10 \ kg$ of gas at a pressure of $10^7 \ N/m^2$. How much gas in $kg$ must be removed so that the final pressure becomes $2.5 \times 10^6 \ N/m^2$? (Assume temperature remains constant.)

At $NTP$,a sample of equal volume of chlorine and oxygen is taken. What is the ratio of the number of molecules?

$A$ balloon contains $1500 \ m^3$ of $He$ at $27^{\circ} C$ and $4 \ atm$ pressure. The volume of $He$ at $-3^{\circ} C$ temperature and $2 \ atm$ pressure will be: (in $m^3$)

The product of the pressure and volume of an ideal gas is

$A$ container $A$ contains an ideal gas at pressure $P$,volume $V$,and temperature $T$. $A$ second container $B$ contains the same gas at pressure $2P$,volume $2V$,and temperature $\frac{T}{2}$. The ratio of the mass of gas in $A$ to that in $B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module