Answer:
B
Explanation:
so you started out with 80
you have to cool it till 60
so it is decreasing which means you have to subtract
Note=if something is increasing or wanting to go up its addition but for this problem 60 is less then 80 so we are doing subtraction
80-60=20
Answer:
0.00011 JK.
The process does NOT violate the second law of thermodynamics
Explanation:
The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.
=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.
The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.
The change in entropy of the cold container = 1/295 = 0.00339 J/K.
Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.
Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.
The process does NOT violate the second law of thermodynamics.
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Answer: 287.8 cm3
Explanation:
Given that:
Initial volume of gas V1 = 350 cm3
Initial pressure of gas P1 = 740 mmHg
New volume V2 = ?
New pressure P2 = 900 mmHg
Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
740 mmHg x 350 cm3 = 900mmHg x V2
V2 = (740 mmHg x 350 cm3) /900mmHg
V2 = 259000 mmHg cm3 / 900mmHg
V2 = 287.8 cm3
Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.
