Question

For a perfect gas, two pressures $P_1$ and $P_2$ are shown in the figure. The graph shows:

• A. $P_1>P_2$
• B. $P_1<P_2$
• C. $P_1 = P_2$
• D. Insufficient data to draw any conclusion

Answer 1

Answer: (A)

$P_1>P_2$

Answer 2

As per ideal gas equation, $V=\frac{nR}{P}T$

$⇒$ Slope of $V-T$ graph is inversely proportional to $P$.

As $m_2 > m_1 ⇒ P_1 > P_2$