Question

A ball has a volume of 6.35 liters and is at a temperature of

Answer 1

Answer:

• Absolute pressure inside the ball: $1.45\; \rm atm$.
• Number of moles of air particles inside the ball, by the ideal gas law: approximately $0.37\; \rm mol$.

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:

$\begin{aligned}& \text{Gauge Pressure} \\ &= \text{Absolute Pressure} - \text{Atomspheric Pressure}\end{aligned}$.

Rearrange to obtain:

$\begin{aligned}& \text{Absolute Pressure} \\ &= \text{Gauge Pressure} + \text{Atomspheric Pressure} \\ &= 0.45\; \rm atm + 1.00 \; \rm atm = 1.45\; \rm atm\end{aligned}$.

Look up the ideal gas constant. This constant comes in a large number of unit combinations. Look for the one that takes $\rm atm$ for pressure and $\rm L$ for volume.

$R = 0.082057\; \rm L \cdot atm \cdot K^{-1}\cdot mol^{-1}$.

Convert the temperature to absolute temperature:

$T = 27.0\; \rm ^\circ C \approx (27.0 + 273.15)\; \rm K = 300.15\; \rm K$.

Assume that the gas inside this ball acts like an ideal gas. Apply the ideal gas law $P \cdot V = n \cdot R \cdot T$ (after rearranging) to find the number of moles of gas particles in this ball:

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &= \frac{1.45\; \rm atm \times 6.35\; \rm L}{0.08205 \; \rm L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300.15\; \rm K} \approx 0.37\; \rm mol\end{aligned}$.

(Rounded to two significant figures, as in the pressure gauge reading.)