3.4 Practice problem answers

Khoi Tran

1. Answers for a), b), c):

   a) P = 10.0 bar

   b) %$_{error}$ = 12.3%

   c) The answer is B.

   • A: False. The pressure inside of the tank will decrease as the gas escapes through the valve.
   • B: True. The total pressure equals to the partial pressure of carbon dioxide as calculated plus the partial pressure of the remaining air.
   • C: False. The van der Waals equation models the behaviour of real gases and accounts for intermolecular attractions. Furthermore, unaccounted intermolecular attractions will cause the actual pressure to decrease, not increase.

2. Using equation (3): $\frac{\text{Rate of effusion of unknown}}{\text{Rate of effusion of } \ce{CO2}} = \frac{\Delta V_{unknown}}{\Delta V_{\ce{CO2}}} = \sqrt{\frac{M_{\ce{CO2}}}{M_{unknown}}}$

$$M_{\text{unknown}} = M_{\ce{CO2}} \cdot (\frac{\Delta V_{\ce{CO2}}}{\Delta V_{\text{unknown}}})^2 = 44.01 \ce{g mol^{-1}} \cdot (\frac{\ce{5.00L - 3.80L}}{\ce{5.00L - 4.11L}}) = 80.0 \ce{g mol^{-1}} = M_{HX}$$

$$\ce{M_X = 80.0 g mol^{-1} - 1.0 g mol^{-1} = 79.0 g mol^{-1} = M_{Br}}$$

3. The answer is B.
   • A: False. When he injects too much helium into the chamber, the pressure increases while the number of moles also increases accordingly. This does not directly affect the R value.
   • B: True. The ideal gas law assumes that gas molecules have negligible volume. However, when they are cramped together under high pressure, these volumes will contribute to the pressure significantly. This causes the PV product to be higher than the ideal nRT.
   • C: False. More significant London dispersion forces will increase intermolecular attractions, which will cause the pressure to be lower than ideal and reduce R value, not increase.
   • D: False. Low temperature will also cause the pressure to decrease proportionally. This does not directly affect the R value.