Answer : The value of work done by an ideal gas is, 37.9 J
Explanation :
Formula used :
Expansion work = External pressure of gas × Volume of gas
Expansion work = 1.50 atm × 0.25 L
Expansion work = 0.375 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 0.375 × 101.3 = 37.9 J
Therefore, the value of work done by an ideal gas is, 37.9 J
Answer:
Option B . heat is taken in
<span>There is a direct correlation between the period number and the energy level for valence electrons. For example, the H and He elements, in period 1, have their outer electrons in the energy level "1". This continues down the rows: all the elements in period 2 have their principal energy level as n = 2, period 3 has n = 3, and so on.</span>
Answer:
P = 0.6815 atm
Explanation:
Pressure = 754 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 754 / 760 atm = 0.9921 atm
Temperature = 294 K
Volume = 3.1 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K
⇒n of helium gas= 0.1274 moles
Surface are = 1257 cm²
For a sphere, Surface area = 4 × π × r² = 1257 cm²
r² = 1257 / 4 × π ≅ 100 cm²
r = 10 cm
The volume of the sphere is :
Where, V is the volume
r is the radius
V = 4190.4762 cm³
1 cm³ = 0.001 L
So, V (max) = 4.19 L
T = 273 K
n = 0.1274 moles
Using ideal gas equation as:
PV=nRT
Applying the equation as:
P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K
<u>P = 0.6815 atm</u>
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Answer:
The amount of salt in the tank at the end of an additional 10 minutes are 4.38lb.
Explanation:
<u>Situation 1:</u>
Tank with 100 gallons of fresh water
<u>Situation 2:</u>
Tank with 100 gallons of fresh water + water with 0.5lb of salt per gallon
After 10 minutes, as the rate in which the new water is poured is 2 gallons per minute, the result is 20 gallons added (2×10=20) . And taking in account that the water contains 0.5 lb of salt per gallon the amount of salt added is 20×0.5= 10lb of salt.
That amount of salt is now in all the water inside the tank which is 100 gallons+ 20 gallons= 120 gallons. <em>That means that in situation 2 we have 10lb of salt in 120 gallons of water.</em>
That mixture is allowed to leave the tank at a rate of 2 gallons per minute so we will have after 10 minutes: 120 gallons- (2×10) gallons= 100 gallons remaining in the tank. And the amount of salt if we remember that we had 10lb in 120 gallons, now in 100 gallons we will have: (100 gallons × 10lb of salt)/ 120 gallons= 8.33 lb of salt.
<u>Situation 3:</u>
Tank with 100 gallons of water with 8.33lb of salt.
After 10 minutes in which fresh water is poured in the tank at a rate of 9 gallons per minute, the result is: 9×10= 90 gallons added to the tank. So now we have 100+90=190 gallons of water in the tank. <em>That means in situation 3 we have 8.33 lb of salt in 190 gallons of water. </em>
That mixture is leaving the tank at a rate of 9 gallons per minute so we have after 10 minutes: (190- (9×10))= 100 gallons of mixture remaining in the tank.
And the amount of salt if we remember that we had 8.33lb in 190 gallons, now in 100 gallons we will have: (100 gallons × 8.33lb of salt)/ 190 gallons= 4.38 lb of salt.