Two bulbs of volumes $1.0 L$ and $2.0 L$ are connected by a tube of negligible volume and together contain $0.60 mol$ of an ideal gas. The smaller bulb is held at $300 K$ and the larger at $400 K$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the common pressure in atm (two decimals) is

Numerical answer type, answer in atm, tolerance 0.02.

Answer: 5.91 atm

The pressure is the same in both bulbs.

n_{total} = \frac{P V _{1}}{R T _{1}} + \frac{P V _{2}}{R T _{2}}

, so

P = \frac{n R}{V _{1} / T _{1} + V _{2} / T _{2}} = \frac{0.60 \times 0.0821}{1.0/300 + 2.0/400} = \frac{0.04926}{0.008333} = 5.911 \approx 5.91 atm