Answer:
The answer for this question depends on the type of meniscus in the cylinder. If it is an upright meniscus like in water, the reading should be taken at the bottom of the meniscus. However if it is an inverted meniscus like in mercury, the reading should be taken at the top of the meniscus.
Answer : The new pressure of the gas will be, 468.66 atm
Explanation :
Boyle's Law : This law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
or,
where,
= initial pressure of the gas = 74 atm
= final pressure of the gas = ?
= initial volume of the gas = 190 ml
= final volume of the gas = 30 ml
Now we put all the given values in the above formula, we get the final or new pressure of the gas.
Therefore, the new pressure of the gas will be, 468.66 atm
A. Archilles ! Hope this helped :)!
The liquid to solid process using the particle theory is as below
- The process that involve change of liquid to solid is known as Freezing
- It involve change from a high energy state to lower energy
- The constant temperature at which a liquid change to solid by giving out heat energy is called freezing point of the liquid
- when liquid are cooled the thermal energy of particles decrease.
- The cohesive forces between the particles strengthen to such extent that particles can have relative motion with each other and they occupy the fixed position, thus liquid is converted to solid
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C