Let's break down your questions one by one, focusing on the concepts of thermodynamics and gas properties. Each part involves some calculations and understanding of specific heat, degrees of freedom, and the relationship between work and heat in thermodynamic processes.
1. Molar Specific Heat of a Gas Mixture
To find the molar specific heat at constant volume (\(C_V\)) of a mixture of gases, we can use the formula:
C_V = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}
In your case, we have:
- Two moles of monoatomic gas (let's assume it's helium, \(C_{V1} = \frac{3}{2}R\))
- Three moles of diatomic gas (like nitrogen, \(C_{V2} = \frac{5}{2}R\))
Now, substituting the values:
- For monoatomic gas: \(n_1 = 2\), \(C_{V1} = \frac{3}{2}R\)
- For diatomic gas: \(n_2 = 3\), \(C_{V2} = \frac{5}{2}R\)
Plugging these into the formula:
C_V = \frac{2 \cdot \frac{3}{2}R + 3 \cdot \frac{5}{2}R}{2 + 3} = \frac{3R + \frac{15}{2}R}{5} = \frac{\frac{21}{2}R}{5} = \frac{21R}{10}
Thus, the molar specific heat of the mixture at constant volume is \(\frac{21R}{10}\).
2. Degrees of Freedom of Gas Molecule
The degrees of freedom of a gas molecule can be determined using the formula for the speed of sound in a gas:
v = \sqrt{\frac{\gamma RT}{M}}
Where:
- \(v\) is the velocity of sound (330 m/s)
- \(\gamma\) is the adiabatic index (ratio of specific heats, \(C_p/C_V\))
- R is the universal gas constant (8.314 J/(mol·K))
- M is the molar mass of the gas in kg/mol
First, we need to find the molar mass (M). Given the density at normal temperature and pressure (NTP) is 1.3 mg/cc, we convert this to kg/m³:
1.3 mg/cc = 1.3 \times 10^{-3} kg/L = 1.3 \times 10^{-3} \times 1000 kg/m³ = 1.3 kg/m³
Using the ideal gas law, \(PV = nRT\), we can find the molar mass:
M = \frac{dRT}{P}
At NTP, \(P = 101325 Pa\) and \(T = 273.15 K\):
M = \frac{1.3 \times 8.314 \times 273.15}{101325} \approx 0.0306 kg/mol
Now, substituting into the speed of sound equation:
330 = \sqrt{\frac{\gamma \cdot 8.314 \cdot 273.15}{0.0306}}
Squaring both sides and solving for \(\gamma\):
108900 = \frac{\gamma \cdot 2270.6}{0.0306}
\(\gamma \approx 1.44\)
For a diatomic gas, \(\gamma = \frac{C_p}{C_V} = \frac{7}{5}\), which implies:
Degrees of freedom = 5
3. Heat Given to the Gas During Isobaric Expansion
In an isobaric process, the heat added to the gas can be calculated using the formula:
Q = W + \Delta U
Where:
- Q is the heat added
- W is the work done (80 J)
- \(\Delta U\) is the change in internal energy
For a diatomic gas, the change in internal energy can be expressed as:
\Delta U = n C_V \Delta T
At constant pressure, we can relate \(\Delta T\) to the work done:
W = P \Delta V = n R \Delta T
From the above, we can express \(\Delta T\) in terms of work:
\Delta T = \frac{W}{nR}
Substituting this back into the equation for \(\Delta U\):
\Delta U = n C_V \left(\frac{W}{nR}\right) = C_V \frac{W}{R}
For a diatomic gas, \(C_V = \frac{5}{2}R\):
\Delta U = \frac{5}{2}R \cdot \frac{80}{R} = 200 J
Now substituting back into the heat equation:
Q = 80 J + 200 J = 280 J
Therefore, the heat given to the gas during this process is 280 J.
