Answer :
Answer:
2.47
Explanation:
Here's how to find the volume of nitrogen gas in the balloon at STP (Standard Temperature and Pressure):
**1. Convert liquid nitrogen volume to mass:**
- We know the volume of liquid nitrogen (V_liquid) = 1.00 mL
- We know the density of liquid nitrogen (ρ_liquid) = 0.807 g/mL
Mass (m_liquid) = Volume (V_liquid) x Density (ρ_liquid)
m_liquid = 1.00 mL * 0.807 g/mL
m_liquid = 0.807 g
**2. Assuming all liquid nitrogen vaporizes:**
The entire mass of liquid nitrogen (0.807 g) will convert to gaseous nitrogen.
**3. Calculate the number of moles of nitrogen gas:**
- Molar mass of nitrogen (M_nitrogen) = 28.02 g/mol (nitrogen molecule has two nitrogen atoms)
Number of moles (n_gas) = Mass (m_liquid) / Molar mass (M_nitrogen)
n_gas = 0.807 g / 28.02 g/mol
n_gas = 0.0288 mol (rounded to four decimals)
**4. Apply the ideal gas law at STP:**
At STP, temperature (T) = 273.15 K and pressure (P) = 1 atm (101.325 kPa).
The ideal gas law equation is: PV = nRT
where:
- P = Pressure
- V = Volume
- n = Number of moles
- R = Ideal gas constant (approximately 8.314 J/mol*K)
- T = Temperature
We want to solve for V (volume of gas). Since we know all other values at STP, we can rearrange the equation:
V = nRT / P
V = 0.0288 mol * 8.314 J/mol*K * 273.15 K / (101.325 kPa)
**Note:** To convert kPa to atm, divide by 101.325.
**5. Convert Joules to Liters*atmosphere (L*atm) for volume:**
1 J/mol*K = 0.00008205 L*atm/mol*K (conversion factor)
V = 0.0288 mol * 8.314 J/mol*K * 273.15 K / (101.325 kPa * 0.00008205 L*atm/mol*K)
V ≈ 2.47 L (rounded to two decimals)
Therefore, the volume of nitrogen gas in the balloon at STP will be approximately 2.47 liters.
