# A 7.35-L container holds a mixture of two gases at 17 °C. The partial pressures of gas A and gas B, respectively, are .415 atm and .720 atm. If .160 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Since pressures are additive, and we know ${P}_{A}$ and ${P}_{B}$, all we need to do is calculate ${P}_{C}$.
${P}_{C} = \frac{{n}_{C} R T}{V}$ $=$ $\frac{0.160 \cancel{\cdot m o l} \times 0.0821 \cancel{\cdot L} \left(\cdot a t m\right) \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 300 \cancel{K}}{7.35 \cancel{L}}$ $=$ ?? $a t m$
P_(TOTAL) = P_C + P_A + P_B=??