(a) The specific heat capacity of carbon dioxide, which has a molecular weight of 44 u, is 849 J kg-1 K-1 at constant pressure. How many degrees of freedom does the carbon dioxide molecule have at room temperature? (b).Calculate the volume occupied by one mole of an ideal gas at a pressure of 92 atm and a temperature of 457 ?C (conditions appropriate to the surface of Venus). Hence estimate the density of the Venusian atmosphere at its surface, assuming a purely carbon dioxide composition. By which factor does this exceed the density of the Earth’s atmosphere at sea level? (c) The NovaLynx 400-8210 pilot balloon has a mass of 10 g and a standard inflated diameter of 46 cm. This balloon is not designed to carry a payload (they’re intended to be tracked from the ground to investigate wind conditions). If the balloon is filled with helium and launched on a day when the pressure is 101 kPa and the temperature is 15 ?C, what is its acceleration as it lifts off?
The Correct Answer and Explanation is:
Let’s solve each part step by step.
(a) Degrees of Freedom of CO₂ at Room Temperature
We are given:
- Specific heat at constant pressure, $c_p = 849 \, \text{J/kg·K}$
- Molecular mass of CO₂, $M = 44 \, \text{g/mol} = 0.044 \, \text{kg/mol}$
We know from kinetic theory:
$$
C_p = \left(1 + \frac{f}{2}\right) R
$$
Where:
- $f$ is the number of degrees of freedom,
- $R = 8.314 \, \text{J/mol·K}$ is the gas constant.
We convert the specific heat per kg to per mole:
$$
C_p^{\text{mol}} = c_p \cdot M = 849 \cdot 0.044 = 37.356 \, \text{J/mol·K}
$$
Now solve for $f$:
$$
37.356 = \left(1 + \frac{f}{2}\right) \cdot 8.314
\Rightarrow \frac{f}{2} = \frac{37.356}{8.314} – 1 = 3.49
\Rightarrow f \approx 7
$$
Answer (a): 7 degrees of freedom — 3 translational + 2 rotational + 2 vibrational (at room temperature, CO₂ exhibits vibrational modes due to being a linear molecule).
(b) Volume and Density on Venus
Given:
- $P = 92 \, \text{atm} = 92 \times 1.013 \times 10^5 = 9.3196 \times 10^6 \, \text{Pa}$
- $T = 457^\circ C = 730 \, \text{K}$
- $n = 1 \, \text{mol}$
- $R = 8.314 \, \text{J/mol·K}$
Using Ideal Gas Law:
$$
PV = nRT \Rightarrow V = \frac{nRT}{P} = \frac{1 \cdot 8.314 \cdot 730}{9.3196 \times 10^6} = 6.51 \times 10^{-4} \, \text{m}^3
$$
Now calculate density $\rho$:
$$
\rho = \frac{\text{mass}}{\text{volume}} = \frac{0.044 \, \text{kg}}{6.51 \times 10^{-4}} \approx 67.6 \, \text{kg/m}^3
$$
Earth’s atmospheric density at sea level ≈ 1.225 kg/m³
$$
\text{Density factor} = \frac{67.6}{1.225} \approx 55.2
$$
Answer (b): Volume ≈ 0.000651 m³, Density ≈ 67.6 kg/m³, about 55 times denser than Earth’s atmosphere.
(c) Acceleration of the Balloon at Lift-off
Given:
- Mass of balloon: $m = 0.01 \, \text{kg}$
- Diameter: $d = 0.46 \, \text{m} \Rightarrow r = 0.23 \, \text{m}$
- Volume: $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.23)^3 \approx 0.051 \, \text{m}^3$
- Helium molar mass ≈ 4 g/mol = 0.004 kg/mol
At $T = 15^\circ C = 288 \, \text{K}$, $P = 101 \times 10^3 \, \text{Pa}$
Using ideal gas law:
$$
n = \frac{PV}{RT} = \frac{101000 \cdot 0.051}{8.314 \cdot 288} \approx 2.15 \, \text{mol}
$$
Mass of helium: $m_{\text{He}} = 2.15 \cdot 0.004 = 0.0086 \, \text{kg}$
Total mass = 0.01 + 0.0086 = 0.0186 kg
Buoyant force: $F_b = \rho_{\text{air}} \cdot V \cdot g = 1.225 \cdot 0.051 \cdot 9.8 \approx 0.612 \, \text{N}$
Weight: $W = m_{\text{total}} \cdot g = 0.0186 \cdot 9.8 \approx 0.182 \, \text{N}$
Net force: $F_{\text{net}} = 0.612 – 0.182 = 0.43 \, \text{N}$
Acceleration:
$$
a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{0.43}{0.0186} \approx 23.1 \, \text{m/s}^2
$$
Answer (c): Acceleration ≈ 23.1 m/s²
Summary:
- (a) CO₂ has 7 degrees of freedom at room temp.
- (b) One mole occupies 0.000651 m³, Venus’s atmosphere is 55× denser than Earth’s.
- (c) Balloon’s lift-off acceleration is ≈ 23.1 m/s²
