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iragen [17]
3 years ago
15

The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens

Chemistry
1 answer:
poizon [28]3 years ago
5 0

Answer:

<u>It increases by a factor of four</u>

Explanation:

Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.

pV = K.......(1)

pV = constant

Charles law : The volume of the gas is directly proportional to temperature at constant pressure.

V = KT

or V/T = K = constant ....(2)

Applying equation (1) and (2)

\frac{PV}{T}=K

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

According to question ,

T2 = 4 (T1)

V2 = V1

Put the value of T2 and V2 , The P2 can be calculated,

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{1}}{4T_{1}}

V1 and V1 cancel each other

T1 and T1 cancel each other

We get,

P1=\frac{P2}{4}

or

P2 = 4 P1

So pressure increased by the factor of four

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A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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