Two identical $1.0 L$ bulbs connected by a narrow tube together contain $0.50 mol$ of an ideal gas. One bulb is held at $300 K$ and the other at $600 K$ . The amount of gas in the colder bulb in mol (three decimals) is

Numerical answer type, answer in mol, tolerance 0.002.

Answer: 0.333 mol

Pressure is common:

n_{1} T_{1} = n_{2} T_{2}

for equal volumes, so

n_{1} / n_{2} = 600/300 = 2

n_{1} = \frac{2}{3} \times 0.50 = 0.333 mol

in the colder bulb.