A sample of argon gas has a volume of 71.3 mL at a pressure of 0.763 atm and a temperature of 37.3 C. What is the final volume of the gas, in mL, when the pressure and temperature are changed to 827 mug and 402 K
The correct answer and explanation is:
To solve this problem, we need to use the Combined Gas Law, which relates the pressure, volume, and temperature of a gas: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
Step 1: Identify known values
- Initial Pressure P1=0.763P_1 = 0.763 atm
- Initial Volume V1=71.3V_1 = 71.3 mL
- Initial Temperature T1=37.3∘C=37.3+273.15=310.45T_1 = 37.3^\circ C = 37.3 + 273.15 = 310.45 K
- Final Pressure P2=827P_2 = 827 μg (This needs to be converted to atm or another gas pressure unit. Here, μg (micrograms) is a unit of mass, not pressure**, so we likely have a unit error here.)
⚠️ Important Note: You wrote “827 μg” for pressure. Micrograms (μg) are not a unit of pressure. Could it have been meant to be 827 mmHg or 827 torr?
Assuming you meant 827 mmHg (which is a common pressure unit), let’s proceed by converting that to atm: P2=827 mmHg760 mmHg/atm=1.088 atmP_2 = \frac{827 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1.088 \text{ atm}
- Final Temperature T2=402T_2 = 402 K
- Final Volume V2=?V_2 = ?
Step 2: Plug into the Combined Gas Law
P1V1T1=P2V2T2⇒V2=P1V1T2P2T1\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \Rightarrow V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} V2=(0.763 atm)(71.3 mL)(402 K)(1.088 atm)(310.45 K)V_2 = \frac{(0.763 \, \text{atm})(71.3 \, \text{mL})(402 \, \text{K})}{(1.088 \, \text{atm})(310.45 \, \text{K})}
Step 3: Calculate
V2=(0.763)(71.3)(402)(1.088)(310.45)=21852.9582337.7236≈64.7 mLV_2 = \frac{(0.763)(71.3)(402)}{(1.088)(310.45)} = \frac{21852.9582}{337.7236} \approx 64.7 \, \text{mL}
✅ Final Answer:
64.7 mL\boxed{64.7 \text{ mL}}
🧠 Explanation (300 words):
This problem involves understanding how a gas behaves under different conditions of pressure and temperature. The Combined Gas Law is a crucial formula derived from Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law. It allows you to calculate the change in volume of a gas when both the temperature and pressure are changing, provided the amount of gas remains constant.
The initial conditions (volume, pressure, and temperature) describe a state of the gas, and we are asked to find the new volume after a change in pressure and temperature. Before using the formula, all units must be consistent: temperatures must be in Kelvin (not Celsius), and pressures must be in the same unit (here, we used atm).
The pressure value “827 μg” initially given is likely a typo for 827 mmHg, which is a valid pressure unit. Since the Combined Gas Law requires consistent units, mmHg was converted to atm using the conversion factor 1 atm = 760 mmHg.
After conversion and substitution into the equation, we solved for the final volume. The result, 64.7 mL, shows that the gas compresses slightly due to increased pressure and a higher temperature, which slightly offsets the pressure effect.
This example shows the importance of careful unit conversion and a deep understanding of how gas particles behave under different thermodynamic conditions.