Question
upstudy study bank question image url

\( 4.3 \quad \begin{array}{l}0,61 \mathrm{~kg} \text { of nitrogen at an absolute pressure of } 100 \mathrm{kPa} \text { and with a volume of } \\ 521 \text { litres has a temperature of } 27^{\circ} \mathrm{C}\end{array} \) Calculate the gas constant for nitrogen. [10] TOTAL: 60

Ask by Simpson Gordon. in South Africa
Mar 15,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The gas constant for nitrogen is approximately \( 285\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \).

Solution

1. Write down the ideal gas law in the form that relates pressure, volume, mass, the specific gas constant, and temperature: \[ pV = mRT \] where \( p \) is the pressure, \( V \) is the volume, \( m \) is the mass, \( R \) is the specific gas constant for nitrogen, and \( T \) is the absolute temperature. 2. Rearranging for \( R \) gives: \[ R = \frac{pV}{mT} \] 3. Convert all given quantities to SI units: - Pressure: \( p = 100\,\text{kPa} = 100\,000\,\text{Pa} \) - Volume: \( V = 521\,\text{L} = \frac{521}{1000}\,\text{m}^3 = 0.521\,\text{m}^3 \) - Temperature: \( T = 27^\circ\text{C} = 27 + 273.15 = 300.15\,\text{K} \) - Mass is already in kilograms: \( m = 0.61\,\text{kg} \) 4. Substitute the converted values into the formula for \( R \): \[ R = \frac{100\,000 \times 0.521}{0.61 \times 300.15} \] 5. Calculate the numerator: \[ 100\,000 \times 0.521 = 52\,100\,\text{Pa}\cdot\text{m}^3 \] 6. Calculate the denominator: \[ 0.61 \times 300.15 \approx 183.0915\,\text{kg}\cdot\text{K} \] 7. Now, compute \( R \): \[ R \approx \frac{52\,100}{183.0915} \approx 284.6\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \] Thus, the gas constant for nitrogen is approximately \[ R \approx 285\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}. \]

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Extra Insights

To calculate the gas constant \( R \) for nitrogen, we use the ideal gas equation, which relates pressure \( P \), volume \( V \), temperature \( T \), and the amount of gas \( n \): \[ PV = nRT \] First, we need to convert the standard conditions to match the units. Here, the pressure is given as \( P = 100 \, \text{kPa} = 100,000 \, \text{Pa} \) and the volume is \( V = 521 \, \text{litres} = 0.521 \, \text{m}^3 \). The temperature must also be converted from Celsius to Kelvin: \[ T = 27^{\circ}C + 273.15 = 300.15 \, K \] Next, we can determine the amount of nitrogen gas \( n \) using its mass. The molecular weight of nitrogen \( N_2 \) is approximately \( 28 \, \text{g/mol} \), or \( 0.028 \, \text{kg/mol} \). Thus, the number of moles \( n \) can be calculated as follows: \[ n = \frac{0.61 \, \text{kg}}{0.028 \, \text{kg/mol}} \approx 21.79 \, \text{mol} \] Now we can rearrange the ideal gas equation to find \( R \): \[ R = \frac{PV}{nT} \] Substituting the values we calculated: \[ R = \frac{(100,000 \, \text{Pa})(0.521 \, \text{m}^3)}{(21.79 \, \text{mol})(300.15 \, \text{K})} \] Calculating \( R \): \[ R \approx \frac{52,100}{6,548.07} \approx 7.95 \, \text{J/(mol K)} \] So, the gas constant for nitrogen in this particular scenario is approximately \( 7.95 \, \text{J/(mol K)} \).

Related Questions

Latest Chemistry Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy