Which of the following statements do you think is the correct one? C. collisions of gas molecules are elastic. Which statement is false? A. The density of a gas is constant as long as its temperature

$$P_1 = 1240 \text{ torr}$$ $$V_1 = 47.2 \text{ L}$$ $$P_2 = 730 \text{ torr}$$ $$V_2 = ?$$

$$V_2 = \frac{P_1V_1}{P_2}$$ $$V_2 = \frac{(1240 \text{ torr})(47.2 \text{ L})}{730 \text{ torr}}$$ $$V_2 = \frac{58528 \text{ L}}{730}$$ $$V_2 \approx 80.175 \text{ L}$$

The closest option is D.

The volume would be approximately D. 80.2 L.

$$\boxed{\text{D. } 80.2 \text{ L}}$$

12. A sample of nitrogen occupies 5.50 L under a pressure of 900 torr at 25°C. At what temperature will it occupy 10.0 L at the same pressure?
This problem involves Charles's Law, which states that for a fixed amount of gas at constant pressure, the volume and absolute temperature are directly proportional ($\frac{V_1}{T_1} = \frac{V_2}{T_2}$).

$$V_1 = 5.50 \text{ L}$$ $$T_1 = 25^\circ\text{C} + 273.15 = 298.15 \text{ K}$$ $$V_2 = 10.0 \text{ L}$$ $$T_2 = ?$$

$$T_2 = \frac{V_2T_1}{V_1}$$ $$T_2 = \frac{(10.0 \text{ L})(298.15 \text{ K})}{5.50 \text{ L}}$$ $$T_2 = \frac{2981.5 \text{ K}}{5.50}$$ $$T_2 \approx 542.09 \text{ K}$$

The closest option is D.

The temperature will be approximately D. 269 °C.

$$\boxed{\text{D. } 269 ^\circ\text{C}}$$

13. Under conditions of fixed temperature and amount of gas, Boyle's law requires that

• D. I, II, and III
  • Boyle's Law states that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional.
  • I. $P_1V_1 = P_2V_2$: This is a common mathematical expression of Boyle's Law.
  • II. $PV = \text{constant}$: This is the fundamental definition of Boyle's Law.
  • III. $P_1/P_2 = V_2/V_1$: This is a rearrangement of $P_1V_1 = P_2V_2$ ($P_1V_1 = P_2V_2 \implies P_1/P_2 = V_2/V_1$).
  • All three statements correctly represent Boyle's Law.

The volume of a sample of nitrogen is 6.00 L at 35°C and 740 torr. What volume will it occupy at STP?
This problem involves the Combined Gas Law, which relates pressure, volume, and temperature for a fixed amount of gas ($\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$). STP (Standard Temperature and Pressure) is $0^\circ\text{C}$ (273.15 K) and 1 atm (760 torr).

$$V_1 = 6.00 \text{ L}$$ $$T_1 = 35^\circ\text{C} + 273.15 = 308.15 \text{ K}$$ $$P_1 = 740 \text{ torr}$$ $$P_2 = 760 \text{ torr (STP)}$$ $$T_2 = 0^\circ\text{C} + 273.15 = 273.15 \text{ K (STP)}$$ $$V_2 = ?$$

$$V_2 = \frac{P_1V_1T_2}{P_2T_1}$$ $$V_2 = \frac{(740 \text{ torr})(6.00 \text{ L})(273.15 \text{ K})}{(760 \text{ torr})(308.15 \text{ K})}$$ $$V_2 = \frac{1213686 \text{ L}}{234194}$$ $$V_2 \approx 5.182 \text{ L}$$

The closest option is D. (Note: The options provided in the image are A. 6.59 L, B. 5.46 L, C. 6.95 L, D. 5.67 L. My calculated value is not among them. Let me recheck the calculation.
$V_2 = \frac{740 \times 6.00 \times 273.15}{760 \times 308.15} = \frac{1213686}{234194} \approx 5.182$ L.
There might be a slight discrepancy with the options or a rounding difference. Let's re-evaluate the options given the calculation. None of the options are close to 5.182 L.
Let's re-check the calculation with slightly different rounding or if there's a common mistake.
$P_1 = 740 \text{ torr}$
$V_1 = 6.00 \text{ L}$
$T_1 = 35^\circ\text{C} = 308.15 \text{ K}$
$P_2 = 760 \text{ torr}$
$T_2 = 0^\circ\text{C} = 273.15 \text{ K}$

$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$
$V_2 = 6.00 \text{ L} \times \frac{740 \text{ torr}}{760 \text{ torr}} \times \frac{273.15 \text{ K}}{308.15 \text{ K}}$
$V_2 = 6.00 \times 0.97368 \times 0.88648$
$V_2 = 5.84208 \times 0.88648$
$V_2 = 5.182 \text{ L}$

The calculation is consistent. Given the options, there might be an error in the question's options or the provided solution (if any was implied by the strike-through on B). If I had to pick the closest, it would be B (5.46 L) or D (5.67 L), but neither is very close to 5.18 L. However, I must provide an answer from the options. Let's assume there's a slight rounding difference or an error in the question's options. I will stick to my calculated value.

Since I must choose from the given options, and my calculated value is approximately 5.18 L, none of the options are very close. However, if we consider the possibility of rounding in the options, 5.46 L (B) and 5.67 L (D) are the closest. Let's re-examine the image for any pre-marked answers. Option B has a strike-through, which might indicate it was considered incorrect by a previous attempt. Option D is 5.67 L.

Let's re-evaluate the calculation one more time to ensure no arithmetic error.
$V_2 = \frac{740 \times 6.00 \times 273.15}{760 \times 308.15} = \frac{1213686}{234194} \approx 5.182 \text{ L}$.
The calculation is robust. Given the options, there's a significant discrepancy. I will state the calculated value and acknowledge the options. If forced to choose, I would pick the closest, but it's not a good fit.

Let's assume there was a typo in the question or options. If the initial volume was slightly different, or the temperature/pressure, it could lead to one of the options.
For example, if $V_2$ was 5.46 L (option B), then $P_1V_1/T_1 = P_2V_2/T_2 \implies V_2 = \frac{740 \times 6.00 \times 273.15}{760 \times 308.15} = 5.182 \text{ L}$.
If $V_2$ was 5.67 L (option D), then $V_2 = 5.182 \text{ L}$.

Given the discrepancy,