For a perfect gas, two pressures $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are shown in figure. The graph shows:

(A) $\mathrm{P}_{1}>\mathrm{P}_{2}$

(B) $\mathrm{P}_{1}<\mathrm{P}_{2}$

(C) $\mathrm{P}_{1}=\mathrm{P}_{2}$

(D) Insufficient data to draw any conclusion

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Heat & Thermodynamics

For a perfect gas, two pressures

P_{1}

and

P_{2}

are shown in figure. The graph shows:

(A) $P_{1} > P_{2}$ ✓ Correct
(B) $P_{1} < P_{2}$
(C) $P_{1} = P_{2}$
(D)Insufficient data to draw any conclusion

Answer: (A)

Solution

From the ideal gas law

P V = n R T

, we get

\frac{V}{T} = \frac{n R}{P}

. The slope of the V-T graph is

\frac{n R}{P}

. A smaller slope means larger pressure. From the figure, the line for

P_{1}

has a smaller slope than

P_{2}

, so

P_{1} > P_{2}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option

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