For a perfect gas,two pressures P_{1} and P_{ | vedclass

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

Question

(A) The ideal gas equation is given by $PV = nRT$.Rearranging this for the volume $V$ as a function of temperature $T$,we get $V = (\frac{nR}{P})T$.Th

Vedclass · Accepted Answer

$P_{1} > P_{2}$

Vedclass · Answer

(A) The ideal gas equation is given by $PV = nRT$.Rearranging this for the volume $V$ as a function of temperature $T$,we get $V = (\frac{nR}{P})T$.This represents a straight line passing through the origin with a slope $m = \frac{nR}{P}$.From the graph,the slope of the line for $P_{2}$ is greater than the slope of the line for $P_{1}$ (since the line for $P_{2}$ is steeper).Therefore,$\frac{nR}{P_{2}} > \frac{nR}{P_{1}}$.This implies that $\frac{1}{P_{2}} > \frac{1}{P_{1}}$,which further simplifies to $P_{1} > P_{2}$.

For a perfect gas,two pressures $P_{1}$ and $P_{2}$ are shown in the figure. The graph shows:

Similar Questions

The volume of a given mass of a gas at $27^{\circ} C$ and $1$ $atm$ is $100$ $cc$. What will be its volume at $327^{\circ} C$ if the pressure remains constant (in $cc$)?

$A$ gas is filled in a cylinder. If its temperature is increased by $20\%$ and its volume decreases by $10\%$, what percentage of the gas will have leaked out?

The following graph represents the $T-V$ curves of an ideal gas (where $T$ is the temperature and $V$ is the volume) at three pressures $P_1, P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. Then the correct relation is:

When a gas filled in a closed vessel is heated by raising the temperature by $1^{\circ} C$,its pressure increases by $0.4 \%$. The initial temperature of the gas is ..........$K$.

The perfect gas equation for $4 \ g$ of hydrogen gas is

Vedclass Test Series

Exam Paper Generator

Online Exam Module