What is the volume of one mole of gas at "Standard Temperature and Pressure" (STP), that is, 1.00 atm and 0°C? Number of mole = 1(n) Temp. T = 0°C = 273.15 K Pressure, P = 1 bar (atm) PV=nRT R = 0.08314 bar 2) What is the volume of one mole of gas at room temperature (20°C)? 3) You inflate your car tires to 2.00 atm pressure at 20°C. After driving on the highway, the temperature of your tires goes up to 50°C. What is the pressure in the tires? 4) What does 1.00 L of nitrogen (N2) weigh at 20°C? Hint: you calculated the volume of one mole of gas at 20°C for question 2. One liter is what fraction of this volume? 5) An internal combustion engine burns fuel inside a cylinder. The heat expands the gases in the cylinder and pushes against the piston. If the temperature in the cylinder increases from 300°C to 600°C, how much does the pressure increase? What if the final temperature is 800°C?
The Correct Answer and Explanation is:
Ideal Gas Calculations and Explanation
To solve these questions, we use the Ideal Gas Law: PV=nRTPV = nRTPV=nRT
Where:
- PPP is pressure
- VVV is volume
- nnn is number of moles
- RRR is the ideal gas constant
- TTT is temperature in Kelvin
We’ll use the gas constant:
- R=0.08314 L\cdotpbar/mol\cdotpKR = 0.08314 \ \text{L·bar/mol·K}R=0.08314 L\cdotpbar/mol\cdotpK
1. Volume of 1 mole of gas at STP (1.00 atm, 0°C)
At standard temperature and pressure:
- P=1.00 barP = 1.00 \ \text{bar}P=1.00 bar
- T=0°C=273.15 KT = 0°C = 273.15 \ \text{K}T=0°C=273.15 K
- n=1.00 moln = 1.00 \ \text{mol}n=1.00 mol
Using the Ideal Gas Law: V=nRTP=(1.00)(0.08314)(273.15)1.00=22.71 LV = \frac{nRT}{P} = \frac{(1.00)(0.08314)(273.15)}{1.00} = 22.71 \ \text{L}V=PnRT=1.00(1.00)(0.08314)(273.15)=22.71 L
Answer: Volume = 22.71 L
2. Volume of 1 mole of gas at 20°C
At:
- T=20°C=293.15 KT = 20°C = 293.15 \ \text{K}T=20°C=293.15 K
- P=1.00 barP = 1.00 \ \text{bar}P=1.00 bar
V=(1.00)(0.08314)(293.15)1.00=24.38 LV = \frac{(1.00)(0.08314)(293.15)}{1.00} = 24.38 \ \text{L}V=1.00(1.00)(0.08314)(293.15)=24.38 L
Answer: Volume = 24.38 L
3. Tire pressure after heating from 20°C to 50°C
Use Gay-Lussac’s Law (constant volume): P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}T1P1=T2P2
Convert to Kelvin:
- T1=20°C=293.15 KT_1 = 20°C = 293.15 \ \text{K}T1=20°C=293.15 K
- T2=50°C=323.15 KT_2 = 50°C = 323.15 \ \text{K}T2=50°C=323.15 K
- P1=2.00 atmP_1 = 2.00 \ \text{atm}P1=2.00 atm
P2=P1⋅T2T1=2.00⋅323.15293.15=2.21 atmP_2 = P_1 \cdot \frac{T_2}{T_1} = 2.00 \cdot \frac{323.15}{293.15} = 2.21 \ \text{atm}P2=P1⋅T1T2=2.00⋅293.15323.15=2.21 atm
Answer: Final pressure = 2.21 atm
4. Mass of 1.00 L of nitrogen gas (N₂) at 20°C
From Q2, 1 mole of gas at 20°C occupies 24.38 L
So, 1.00 L is: 1.0024.38=0.041 mol\frac{1.00}{24.38} = 0.041 \ \text{mol}24.381.00=0.041 mol
Molar mass of nitrogen (N₂) = 28.02 g/mol
Mass: 0.041⋅28.02=1.15 g0.041 \cdot 28.02 = 1.15 \ \text{g}0.041⋅28.02=1.15 g
Answer: 1.15 g
5. Pressure increase in engine cylinder: 300°C to 600°C and 800°C
Convert to Kelvin:
- Initial: T1=300+273.15=573.15 KT_1 = 300 + 273.15 = 573.15 \ \text{K}T1=300+273.15=573.15 K
- Case 1 Final: T2=600+273.15=873.15 KT_2 = 600 + 273.15 = 873.15 \ \text{K}T2=600+273.15=873.15 K
- Case 2 Final: T3=800+273.15=1073.15 KT_3 = 800 + 273.15 = 1073.15 \ \text{K}T3=800+273.15=1073.15 K
Again using: P2P1=T2T1\frac{P_2}{P_1} = \frac{T_2}{T_1}P1P2=T1T2
Case 1: P2P1=873.15573.15=1.52\frac{P_2}{P_1} = \frac{873.15}{573.15} = 1.52P1P2=573.15873.15=1.52
Case 2: P3P1=1073.15573.15=1.87\frac{P_3}{P_1} = \frac{1073.15}{573.15} = 1.87P1P3=573.151073.15=1.87
Answers:
- Pressure increases by 1.52× when temperature rises to 600°C
- Pressure increases by 1.87× when temperature rises to 800°C
Summary
This set of questions uses the Ideal Gas Law and gas behavior laws (Boyle’s, Charles’s, Gay-Lussac’s) to relate pressure, volume, and temperature. Gas volume expands with temperature, and pressure increases if gas is heated in a fixed volume. These principles are critical in both everyday and industrial applications like tire inflation and internal combustion engines.
