Answer:
894 deg K
Explanation:
The computation is shown below:
Given that
V1 denotes the initial volume of gas = 2.00 L
T1 denotes the initial temperature of gas = 25 + 273 = 298 K
V2 denotes the final volume of gas = 6.00 L
T2 = ?
Based on the above information
Here we assume that the pressure is remain constant,
So,
V1 ÷ T1 = V2 ÷ T2
T2 = T1 × V2 ÷ V1
= (298)(6) ÷ (2)
= 894 deg K
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Answer:
108 kPa
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
Data:
p₁ = ?; V₁ = 34.3 L; T₁ = 31.5 °C
p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C
Calculations:
(a) Convert temperatures to <em>kelvins
</em>
T₁ = (31.5 + 273.15) K = 304.65 K
T₂ = (21.0 + 273.15) K = 294.15 K
(b) Calculate the <em>pressure
</em>
p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)
= 122.2 kPa × 0.8542 × 1.0357
= 108 kPa
