A sample of oxygen occupies a volume of 160 dm3 at 91° C. What will be volume of oxygen when the temperature drops to 0.00° C?

Question

Natalie Anton · Accepted Answer

The volume of the oxygen will be 120 dm³.

This is an example of Charles' Law.

#V_1/T_1 = V_2/T_2#

#V_1# = 160 dm³; #T_1# = (91 + 273.15) K = 364 K
#V_2# = ?; #T_2# = (0.00 + 273.15) K = 273.15 K

#V_2 = V_1× T_2/T_1# = 160 dm³ ×#(273.15"K")/(364"K")# = 120 dm³

This makes sense. The final Kelvin temperature was about 25 % lower than the initial temperature. The final volume should be about 25 % (40 dm³) less than the initial volume.

Avery Adams · Answer

undefined To find the final volume of oxygen when the temperature drops to 0.00°C, we can use the combined gas law, which relates the initial and final volumes and temperatures of a gas sample:

$$ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} $$

Given:
- Initial volume, $V_1 = 160 \, \text{dm}^3$
- Initial temperature, $T_1 = 91 \, \text{°C}$ (convert to Kelvin by adding 273.15: $T_1 = 91 + 273.15 \, \text{K}$)
- Final temperature, $T_2 = 0.00 \, \text{°C}$ (convert to Kelvin: $T_2 = 0.00 + 273.15 \, \text{K}$)
- We are solving for the final volume, $V_2$

Rearranging the equation to solve for $V_2$:

$$ V_2 = \frac{T_2 \times P_1 \times V_1}{T_1 \times P_2} $$

Since the pressure (P) is constant in this case, it cancels out:

$$ V_2 = \frac{T_2 \times V_1}{T_1} $$

Now, plug in the values to calculate $V_2$.