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Answer:
The value of the heat capacity of the Calorimeter = 54.4
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water = 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q = ΔT + ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 + × 11.24
611.37 = × 11.24
= 54.4
This is the value of the heat capacity of the Calorimeter.
The volume at 100 mmHg : 0.656 L
<h3>
Further explanation</h3>
Boyle's Law
<em>At a constant temperature, the gas volume is inversely proportional to the pressure applied </em>
V₁=3.5 L
P₁=2.5 kPa=18,7515 mmHg
P₂=100 mmHg
<h3>
Answer:</h3>
78.75 K
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial pressure, P₁ = 500 torr
- Initial temperature,T₁ = 225 K
- Initial volume, V₁ = 3.3 L
- Final volume, V₂ = 2.75 L
- Final pressure, P₂ = 210 torr
We are required to calculate the new temperature, T₂
- To find the new temperature, T₂ we are going to use the combined gas law;
- According to the combined gas law;
P₁V₁/T₁ = P₂V₂/T₂
We can calculate the new temperature, T₂;
Rearranging the formula;
T₂ =(P₂V₂T₁) ÷ (P₁V₁)
= (210 torr × 2.75 L × 225 K) ÷ (500 torr × 3.3 L)
= 78.75 K
Therefore, the new volume of the sample is 78.75 K