The final volume of the methane gas in the container is 6.67 L.
The given parameters;
- <em>initial volume of gas in the container, V₁ = 2.65 L</em>
- <em>initial number of moles of gas, n₁ = 0.12 mol</em>
- <em>additional concentration, n = 0.182 mol</em>
The total number of moles of gas in the container is calculated as follows;
The final volume of gas in the container is calculated as follows;
Thus, the final volume of the methane gas in the container is 6.67 L.
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Answer:
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Answer:
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Answer:
- 278.85 J
Explanation:
Given that:
Pressure = 1.1 atm
The initial volume V₁ = 0.0 L
The final volume V₂ = 2.5 L
The work that takes place in a reaction at constant pressure can be expressed by using the equation:
W = P(V₂ - V₁ )
Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:
W = - P(V₂ - V₁ )
W = -1.1 atm ( 2.5 - 0.0) L
W = -1.1 atm (2.5 L)
W = -2.75 atm L
Recall that:
1 atm L = 101.4 J
Therefore;
-2.75 atm L = ( -2.75 × 101.4 )J
= -278.85 J
Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J
