Question

1. A gas expands adiabatically, and its volume doubles, while its absolute temperature drops 1.32 times. What is the number of degrees of freedom the gas molecules have?

2. Two identical systems each contain ν = 0.1 mole of an ideal gas at T = 300 K and p = 2.0 × 105 Pa. The pressure in the two systems is reduced by a factor 2, allowing the systems to expand, one adiabatically and one isothermally. What are the final temperatures and volumes of each system? Assume that γ = 5/3.

Answer 1

Answer on Question #74025-Physics-Molecular Physics-Thermodynamics

1. A gas expands adiabatically, and its volume doubles, while its absolute temperature drops 1.32 times. What is the number of degrees of freedom the gas molecules have?

Solution

$$TV^{\gamma - 1} = const$$ $$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$ $$\frac{1}{1.32} = \left(\frac{1}{2}\right)^{\gamma - 1}$$ $$2^{\gamma - 1} = 1.32$$ $$\gamma = 1 + \log_2 1.32 = 1.4 = \frac{7}{5}$$

Thus, it is diatomic gas. Therefore, the gas molecules have 5 degrees of freedom.

2. Two identical systems each contain $\nu = 0.1$ mole of an ideal gas at $T = 300\,\mathrm{K}$ and $p = 2.0 \times 10^5\,\mathrm{Pa}$. The pressure in the two systems is reduced by a factor 2, allowing the systems to expand, one adiabatically and one isothermally. What are the final temperatures and volumes of each system? Assume that $\gamma = 5/3$.

Solution

1) Adiabatic

$$pv = vRT$$ $$\frac{p}{2}v = vRT_f$$ $$p^{1 - \gamma}T^{\gamma} = const \rightarrow T_f = T\left(\frac{1}{2}\right)^{\frac{\gamma - 1}{\gamma}} = 300\left(\frac{1}{2}\right)^{\frac{5/3 - 1}{5/3}} = 227\,\mathrm{K}$$ $$v = \frac{2vRT_f}{p} = \frac{2(0.1)(8.31)(227)}{2.0 \cdot 10^5} = 0.0019\,\mathrm{m}^3.$$

2) Isothermal

$$pv = vRT$$ $$\frac{p}{2}v = vRT$$ $$T = const \rightarrow T_f = 300\,\mathrm{K}$$ $$v = \frac{2vRT}{p} = \frac{2(0.1)(8.31)(300)}{2.0 \cdot 10^5} = 0.0025\,\mathrm{m}^3.$$

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