A $2.0 L$ sample of an ideal gas at $300 K$ and $1.0 atm$ and a $3.0 L$ sample of the same gas at $400 K$ and $2.0 atm$ are transferred together into an evacuated $5.0 L$ vessel held at $350 K$ . The final pressure in atm (two decimals) is

Numerical answer type, answer in atm, tolerance 0.02.

Answer: 1.52 atm

Moles:

n_{1} = \frac{1.0 \times 2.0}{R \times 300}

n_{2} = \frac{2.0 \times 3.0}{R \times 400}

P = \frac{( n _{1} + n _{2} ) R \times 350}{5.0} = \frac{350}{5.0} (\frac{2.0}{300} + \frac{6.0}{400}) = 70 \times 0.02167 = 1.517 \approx 1.52 atm

(

R

cancels).