ΔG > 0
is always true for the freezing of water.
Explanation:
- The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
- The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
- Entropy will decline elsewhere, provided it increases by at least as much elsewhere.
Answer:
4477381.7 calories/pound
Explanation:
It is given that,
When a candle burns it produced 41,300 Joules per 1 gram.
We need to convert it into calories per pound.
We know that,
1 cal = 4.184 J
⇒ 1 J = (1/4.184) cal
1 pound = 453.592 grams
⇒1 g = (1/453.592) pounds
Now,
Hence, 41,300 Joules/gram = 4477381.7 calories/pound.
Answer:
χH₂ = 0.4946
χN₂ = 0.4130
χAr = 0.0923
Explanation:
The total pressure of the mixture (P) is:
P = pH₂ + pN₂ + pAr
P = 443.0 Torr + 369.9 Torr + 82.7 Torr
P = 895.6 Torr
We can find the mole fraction of each gas (χ) using the following expression.
χi = pi / P
χH₂ = pH₂ / P = 443.0 Torr/895.6 Torr = 0.4946
χN₂ = pN₂ / P = 369.9 Torr/895.6 Torr = 0.4130
χAr = pAr / P = 82.7 Torr/895.6 Torr = 0.0923
It is a scientific hypothesis. A scientific hypothesis must be testable, however there is a significantly more grounded necessity that a testable speculation must meet before it can truly be viewed as logical. This foundation comes essentially from crafted by the rationalist of science Karl Popper, and is called "falsifiability".
Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is 
.
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas = 
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is .
