Answer:
Explanation:
given,
In first case Volume remains constant.
Work done in the first case is zero.
In Second case Volume change
V₁ = 0.2 m³
V₂ = 0.11 m³
Pressure, P = 5.5 x 10⁵ Pa
Work done = Pressure x change in volume
W = P ΔV
Hence, Work done when volume changes is equal to
Explanation:
Unclear question. The clear rendering reads;
"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?
Givens
=====
V
= 4.00 L
T
= 273oK We're assuming the temperature does not change, just the
pressure.
n
= 0.864 moles
R
= 8.314 joules / mole * oK
P
= ?????
Formula
======
PV
= n*R*T
P
= n*R*T/V
P
= 0.864 * 8.314 * 273 / 4
P
= 490 kpa
You
have to add 1.6 – 0.864 = 0.736 moles of gas.
We
have to assume that the temperature and pressure remain the same when
we add the 0.736 moles of gas. We are now looking for the volume.
PV
= n*R*T
<span>
V
= 0.736 * 8.314 * 273 / 490</span>
V
= 3.41 L Remember this is at about 4 atmospheres so we have to
convert to Standard Pressure.
Total
Volume = 3.41 + 4.00 = 4.41
V1
* P1 = V2 * P2
P1
= 490 kPa
P2
= 101 kPa
V1
= 7.41 L
V2
= ????
<span>
<span>
7.41*
490 = V2 * 101
V2
= 7.41 * 490 / 101
V2
= 35.94 L
</span>
</span>
<span>You
had 4 L now you need 31.94 more.</span>
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