<span>We know that the relation between pressure, volume and temperature is constant for ideal gases, so then
P1*V1/T1 = P2*V2/T2
As the volume is held constant, we will have:
P1/T1 = P2/T2
As the pressure is doubled and the temperature in kelvins is 211.15 K, then:
P1/211.15 = 2*P1/T2
T2 = 2*211.15 = 422.3 K
T2 = 422.3 - 273.15 = 149.15
So the final temperature reaches 149.15 ÂşC</span>
Water vapour and a drop in temperature to condense water vapour into a cloud. Something to see the cloud also, such as "pollution" particles.
The charge will most likely leave the electric field near C) Y
Answer: Yes both gases would have the same entropy.
Explanation:
The formula for the change in the entropy is as follows,
Here, \Delta S is the change in the entropy, Q is the heat transfer and T is the temperature.
If the temperature of the system increases then there will be increase in the entropy as the randomness of the system increases.
In the given problem, if the both gases were initially at the same absolute temperature. Then there will be same entropy change in both gases.
Therefore, yes both gases would have the same entropy.
Answer:
= 8.46 × 10³N
Explanation:
W = 1.3 × 10⁴
v₁ = 0
v (friction) = 95km/h = 26.4 m/s
t = 5s
f = 1450N
mass of body = W / g
=1.3 × 10⁴ / 9.8
=1327 kg
v(friction) = v + at
substitute in eqn 1
26.4 = 5a
a = 5.28 m/s
Force Net = F₁ + F₂ + F₃ = ma
F (engine) = ma + F
= 1327 (5.28) + 1450
= 8.46 × 10³N