-32.0 °C, 4.96 L sample of mustard gas at 0.994 atm of pressure. What is the pressure, in atm, of that gas if compressed 1.92 L?
回答 (2)
2016-11-21 4:50 pm
The temperature is kept unchanged.
Before compression: P₁ = 0.994 atm, V₁ = 4.96 L
After compression: P₂ = ? atm, V₂ = 1.92 L
For a fixed amount of gas at constant temperature (Boyle’s Law):
P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
Final pressure, P₂ = (0.994 atm) × (4.96/1.92) = 2.57 atm
Before compression: P₁ = 0.994 atm, V₁ = 4.96 L
After compression: P₂ = ? atm, V₂ = 1.92 L
For a fixed amount of gas at constant temperature (Boyle’s Law):
P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
Final pressure, P₂ = (0.994 atm) × (4.96/1.92) = 2.57 atm
2016-11-21 4:52 pm
Mustard gas at -32.0 °C is a solid. It doesn't compress.
If you ignore that, and suppose we're talking about an ideal gas, and supposing the temperature doesn't change:
(0.994 atm) x (4.96 L / 1.92 L) = 2.57 atm
If you ignore that, and suppose we're talking about an ideal gas, and supposing the temperature doesn't change:
(0.994 atm) x (4.96 L / 1.92 L) = 2.57 atm
收錄日期: 2021-04-18 15:50:48
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