Answer on Question #42963-Physics-Molecular Physics-Thermodynamics
A 600 g of helium gas occupy 2 liters at 80 c suddenly expanded (adiabatic) to volume 3 liter calculate
m = 600 g = 0.6 k g , V 1 = 2 ⋅ 1 0 − 3 m 3 , V 2 = 3 ⋅ 1 0 − 3 m 3 , T = ( 80 + 273 ) K = 353 K . m = 600g = 0.6\,kg, V_1 = 2 \cdot 10^{-3}m^3, V_2 = 3 \cdot 10^{-3}m^3, T = (80 + 273)K = 353\,K. m = 600 g = 0.6 k g , V 1 = 2 ⋅ 1 0 − 3 m 3 , V 2 = 3 ⋅ 1 0 − 3 m 3 , T = ( 80 + 273 ) K = 353 K .
1-molar heat capacities at constant volume and at constant pressure and gamma constant
Solution
Helium is a monoatomic gas. Its molar heat capacities at constant volume is
c V = 3 2 R = 12.5 J m o l K . c_V = \frac{3}{2}R = 12.5\,\frac{J}{mol\,K}. c V = 2 3 R = 12.5 m o l K J .
Its molar heat capacities at constant pressure is
c p = c V + R = 5 2 R = 20.8 J m o l K . c_p = c_V + R = \frac{5}{2}R = 20.8\,\frac{J}{mol\,K}. c p = c V + R = 2 5 R = 20.8 m o l K J .
Its gamma constant is
γ = c p c V = 5 2 R 3 2 R = 5 3 ≈ 1.67. \gamma = \frac{c_p}{c_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \approx 1.67. γ = c V c p = 2 3 R 2 5 R = 3 5 ≈ 1.67.
2-intial pressure
Solution
From the ideal gas law ( p V = m M R T ) (pV = \frac{m}{M} RT) ( p V = M m RT )
p 1 V 1 = m M R T 1 → p 1 = m R T 1 M V 1 = 0.6 k g ⋅ 8.31 J m o l K ⋅ 353 K 4 ⋅ 1 0 − 3 kg mol ⋅ 2 ⋅ 1 0 − 3 m 3 = 2.2 ⋅ 1 0 8 P a . p_1 V_1 = \frac{m}{M} RT_1 \rightarrow p_1 = \frac{m RT_1}{M V_1} = \frac{0.6\,kg \cdot 8.31\,\frac{J}{mol\,K} \cdot 353\,K}{4 \cdot 10^{-3}\frac{\text{kg}}{\text{mol}} \cdot 2 \cdot 10^{-3}m^3} = 2.2 \cdot 10^8 Pa. p 1 V 1 = M m R T 1 → p 1 = M V 1 m R T 1 = 4 ⋅ 1 0 − 3 mol kg ⋅ 2 ⋅ 1 0 − 3 m 3 0.6 k g ⋅ 8.31 m o l K J ⋅ 353 K = 2.2 ⋅ 1 0 8 P a .
3-final temperature and pressure
Solution
For adiabatic process
p 1 V 1 γ = p 2 V 2 γ → p 2 = p 1 ( V 1 V 2 ) γ = 2.2 ⋅ 1 0 8 P a ( 2 ⋅ 1 0 − 3 m 3 3 ⋅ 1 0 − 3 m 3 ) 5 3 = 1.1 ⋅ 1 0 8 P a . p_1 V_1^\gamma = p_2 V_2^\gamma \rightarrow p_2 = p_1 \left(\frac{V_1}{V_2}\right)^\gamma = 2.2 \cdot 10^8 Pa \left(\frac{2 \cdot 10^{-3}m^3}{3 \cdot 10^{-3}m^3}\right)^{\frac{5}{3}} = 1.1 \cdot 10^8 Pa. p 1 V 1 γ = p 2 V 2 γ → p 2 = p 1 ( V 2 V 1 ) γ = 2.2 ⋅ 1 0 8 P a ( 3 ⋅ 1 0 − 3 m 3 2 ⋅ 1 0 − 3 m 3 ) 3 5 = 1.1 ⋅ 1 0 8 P a .
From the ideal gas law ( p V = m M R T ) (pV = \frac{m}{M} RT) ( p V = M m RT )
T 2 = p 2 V 2 p 1 V 1 T 1 = 1 2 ⋅ 3 2 ⋅ 353 K = 265 K . T_2 = \frac{p_2 V_2}{p_1 V_1} T_1 = \frac{1}{2} \cdot \frac{3}{2} \cdot 353\,K = 265\,K. T 2 = p 1 V 1 p 2 V 2 T 1 = 2 1 ⋅ 2 3 ⋅ 353 K = 265 K .
4-change in root mean square velocity
Solution
The root mean square velocity is v r m s = 3 k T m v_{rms} = \sqrt{\frac{3kT}{m}} v r m s = m 3 k T
Δ v r m s = 3 R T 2 M − 3 R T 1 M = 3 R T 1 M ( T 2 T 1 − 1 ) = 3 ⋅ 8.31 J m o l K ⋅ 353 K 4 ⋅ 1 0 − 3 k g m o l ( 3 4 − 1 ) = − 199 m s . \Delta v_{rms} = \sqrt{\frac{3RT_2}{M}} - \sqrt{\frac{3RT_1}{M}} = \sqrt{\frac{3RT_1}{M}} \left( \sqrt{\frac{T_2}{T_1}} - 1 \right) = \sqrt{ \frac{3 \cdot 8.31 \frac{J}{molK} \cdot 353 K }{4 \cdot 10^{-3} \frac{kg}{mol}} } \left( \sqrt{ \frac{3}{4} } - 1 \right) = -199 \frac{m}{s}. Δ v r m s = M 3 R T 2 − M 3 R T 1 = M 3 R T 1 ( T 1 T 2 − 1 ) = 4 ⋅ 1 0 − 3 m o l k g 3 ⋅ 8.31 m o l K J ⋅ 353 K ( 4 3 − 1 ) = − 199 s m .
5-change in mean kinetic energy of molecule
Solution
The mean kinetic energy of molecule is 3 2 k T \frac{3}{2} kT 2 3 k T
3 2 k Δ T = 3 2 ⋅ 1.38 ⋅ 1 0 − 23 J K ( 265 − 353 ) K = − 1.82 ⋅ 1 0 − 21 J . \frac{3}{2} k \Delta T = \frac{3}{2} \cdot 1.38 \cdot 10^{-23} \frac{J}{K} (265 - 353)K = -1.82 \cdot 10^{-21} J. 2 3 k Δ T = 2 3 ⋅ 1.38 ⋅ 1 0 − 23 K J ( 265 − 353 ) K = − 1.82 ⋅ 1 0 − 21 J .
6-change in molar kinetic energy
Solution
The molar kinetic energy is 3 2 R T \frac{3}{2} RT 2 3 RT
3 2 R Δ T = c V Δ T = 12.5 J m o l K ⋅ ( 265 − 353 ) K = − 1.1 k J m o l . \frac{3}{2} R \Delta T = c_V \Delta T = 12.5 \frac{J}{molK} \cdot (265 - 353)K = -1.1 \frac{kJ}{mol}. 2 3 R Δ T = c V Δ T = 12.5 m o l K J ⋅ ( 265 − 353 ) K = − 1.1 m o l k J .
7-change in internal energy and its sign
Solution
Δ U = m M c V Δ T = 0.6 k g 4 ⋅ 1 0 − 3 k g m o l ⋅ 12.5 J m o l K ⋅ ( 265 − 353 ) K = − 165 k J . \Delta U = \frac{m}{M} c_V \Delta T = \frac{0.6 k g}{4 \cdot 10^{-3} \frac{kg}{mol}} \cdot 12.5 \frac{J}{molK} \cdot (265 - 353)K = -165 kJ. Δ U = M m c V Δ T = 4 ⋅ 1 0 − 3 m o l k g 0.6 k g ⋅ 12.5 m o l K J ⋅ ( 265 − 353 ) K = − 165 k J .
8-the mechanical work
Solution
For adiabatic process work done by gas is
W = − Δ U = 165 k J . W = -\Delta U = 165 kJ. W = − Δ U = 165 k J .
9-what does this work?
Answer
W W W
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#339914 on Dec 2023