a. W = 0 J
b. W = - 308.028 J
<h3>Further explanation</h3>
Given
Nitrogen gas expands in volume from 1.6 L to 5.4 L
Required
The work done
Solution
Isothermal :
W = -P . ΔV
Input the value :
a. At a vacuum, P = 0
So W = 0
b. At pressure = 0.8 atm
W = - 0.8 x ( 5.4 - 1.6)
W = -3.04 L.atm ( 1 L.atm = 101.325 J)
W = - 3.04 x 101.325
W = - 308.028 J
Answer:
Saturated solution = 180 gram
Explanation:
Given:
Solubility of Z = 60 g / 100 g water
Given temperature = 20°C
Amount of water = 300 grams
Find:
Saturated solution
Computation:
Saturated solution = [Solubility of Z] × Amount of water
Saturated solution = [60 g / 100 g] × 300 grams
Saturated solution = [0.6] × 300 grams
Saturated solution = 180 gram
Organisms is an environment or habitat for species
<span>1. </span>To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
V2 = 104.1 x 478 / 88.2
<span> V2 =564.17 cm^3</span>
Answer:
ΔS° = - 47.2 J/mol.K
Explanation:
ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10
∴ S°mH2O(l) = 69.9 J/mol.K
∴ S°mP4O10 = 231 J/mol.K
∴ S°mH3PO4 = 150.8 J/mol.K
⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231
⇒ ΔS° = - 47.2 J/mol.K
