Answer:
- Absolute pressure inside the ball: [tex]1.45\; \rm atm[/tex].
- Number of moles of air particles inside the ball, by the ideal gas law: approximately [tex]0.37\; \rm mol[/tex].
Explanation:
The gauge pressure inside the ball gives the absolute pressure inside the ball, relative to the atmospheric pressure outside the ball. In other words:
[tex]\begin{aligned}& \text{Gauge Pressure} \\ &= \text{Absolute Pressure} - \text{Atomspheric Pressure}\end{aligned}[/tex].
Rearrange to obtain:
[tex]\begin{aligned}& \text{Absolute Pressure} \\ &= \text{Gauge Pressure} + \text{Atomspheric Pressure} \\ &= 0.45\; \rm atm + 1.00 \; \rm atm = 1.45\; \rm atm\end{aligned}[/tex].
Look up the ideal gas constant. This constant comes in a large number of unit combinations. Look for the one that takes [tex]\rm atm[/tex] for pressure and [tex]\rm L[/tex] for volume.
[tex]R = 0.082057\; \rm L \cdot atm \cdot K^{-1}\cdot mol^{-1}[/tex].
Convert the temperature to absolute temperature:
[tex]T = 27.0\; \rm ^\circ C \approx (27.0 + 273.15)\; \rm K = 300.15\; \rm K[/tex].
Assume that the gas inside this ball acts like an ideal gas. Apply the ideal gas law [tex]P \cdot V = n \cdot R \cdot T[/tex] (after rearranging) to find the number of moles of gas particles in this ball:
[tex]\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &= \frac{1.45\; \rm atm \times 6.35\; \rm L}{0.08205 \; \rm L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300.15\; \rm K} \approx 0.37\; \rm mol\end{aligned}[/tex].
(Rounded to two significant figures, as in the pressure gauge reading.)
