The figure shows the isotherms of a fixed mas | vedclass

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

Question

(B) From the ideal gas equation,$PV = nRT$,we can write:$\frac{1}{V} = \frac{P}{nRT}$For a fixed mass of gas,$n$ is constant. Thus,$\frac{1}{V} = \lef

Vedclass · Accepted Answer

$T_A < T_B < T_C$

Vedclass · Answer

(B) From the ideal gas equation,$PV = nRT$,we can write:$\frac{1}{V} = \frac{P}{nRT}$For a fixed mass of gas,$n$ is constant. Thus,$\frac{1}{V} = \left(\frac{1}{nRT}\right)P$.This represents a straight line passing through the origin with a slope $m = \frac{1}{nRT}$.Since the slope $m$ is inversely proportional to temperature $T$ $(m \propto \frac{1}{T})$,the line with the largest slope corresponds to the lowest temperature.Looking at the graph,the slope of line $A$ is the greatest,followed by $B$,and then $C$ (i.e.,$m_A > m_B > m_C$).Therefore,the temperatures must follow the order $T_A < T_B < T_C$.

The figure shows the isotherms of a fixed mass of an ideal gas at three temperatures $T_A, T_B$ and $T_C$. Then:

Similar Questions

$A$ gas at $27^{\circ}C$ has a volume $V$ and pressure $P$. On heating,its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be ...... $^{\circ}C$.

$A$ pressure $P-$absolute temperature $T$ diagram was obtained when a given mass of gas was heated. During the heating process from state $1$ to state $2$,the volume:

State Avogadro's law for matter.

When the pressure of a gas is increased by $0.4\%$,its temperature increases by $1\,^{\circ}\text{C}$. What is the initial temperature of the gas?

Two vessels $A$ and $B$ are of the same size and are at the same temperature. Vessel $A$ contains $1 \ g$ of hydrogen and vessel $B$ contains $1 \ g$ of oxygen. If $P_{A}$ and $P_{B}$ are the pressures of the gases in $A$ and $B$ respectively,then the ratio $\frac{P_{A}}{P_{B}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module