Question

A box containing N molecules of a perfect gas at temperature $T_{1}$ and pressure $P_{1}$. The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is $P_{2}$ and temperature $T_{2}$, then

• A. $P_{2} = P_{1}$, $T_{2} = T_{1}$
• B. $P_{2} = P_{1}$, $T_{2} = \frac{T_{1}}{2}$
• C. $P_{2} = 2P_{1}$, $T_{2} = T_{1}$
• D. $P_{2} = 2P_{1}$, $T_{2} = \frac{T_{1}}{2}$

Answer 1

Answer: (B)

$P_{2} = P_{1}$, $T_{2} = \frac{T_{1}}{2}$

Answer 2

Kinetic energy of N molecule of gas $E = \frac{3}{2}NkT$

Initially $E_{1} = \frac{3}{2}N_{1}kT_{1}$ and finally $E_{2} = \frac{3}{2}N_{2}kT_{2}$

But according to problem $E_{1} = E_{2}$ and $N_{2} = 2N_{1}$

∴ $\frac{3}{2}N_{1}kT_{1} = \frac{3}{2}(2N_{1})kT_{2}$ ⇒ $T_{2} = \frac{T_{1}}{2}$

Since the kinetic energy constant $\frac{3}{2}N_{1}kT_{1} = \frac{3}{2}N_{2}kT_{2}$

⇒ $N_{1}T_{1} = N_{2}T_{2}$ ∴ NT = constant

From ideal gas equation of N molecule $PV = NkT$

⇒ $P_{1}V_{1} = P_{2}V_{2}$ ∴ $P_{1} = P_{2}$

$\lbrack\text{As }V_{1} = V_{2}$ and NT = constant]