A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
1 answer:
Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13
From ideal gas equation P V = m RT
⇒ P =
Put all the values in above formula we get
⇒ P = × 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
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Answer:
The body temperature would rise by 47.85 °C
The amount of water the body evaporates is 4.15 kg.
This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature
Explanation:
Considering the relationship (between the heat released and the mass of the object) as shown below
q = msΔT
where q is the heat released per day =
m is the mass of the body = 50 kg
ΔT is the temperature rise = ?
s is the specific heat of water =
substituting values we have
=
ΔT
ΔT =
= 47.85°C
To maintain the normal body temperature (98.6F = 37°C) the amount of heat released by metabolism activity must be utilized for evaporation of some amount of water
Hence
Note (1 kg = 1000 g)
This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature