Answer:
319.15^{o}C[/tex]
Explanation:
When all other variables are constant, we are allowed to use the formula
![\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to KelvinNow to calculate T2[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%20%7D%7BV_%7B2%7D%20%7D%20%3D%20%5Cfrac%7BT_%7B1%7D%20%7D%7BV_%7B1%7D%20%7D%20%5C%5C%3C%2Fp%3E%3Cp%3EWhich%20can%20be%20rewritten%20as%20T_%7B2%7D%20%3D%20%5Cfrac%7BT_%7B1%7D%20V_%7B2%7D%20%7D%7BV_%7B1%7D%20%7D%3C%2Fp%3E%3Cp%3Eif%20you%20make%20T2%20the%20subject%20of%20the%20formula.%20This%20formula%20is%20true%20only%20if%20temperature%20is%20in%20Kelvin%20not%20degrees%20Celsius%20so%20T1%20must%20be%20converted%20to%20Kelvin%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ENow%20to%20calculate%20T2%3C%2Fp%3E%3Cp%3E%5Btex%5DT_%7B2%7D%3D%20%5Cfrac%7B296.15K%2A3.38.10%5E%7B3%7DL%20%7D%7B1.69.10%5E%7B3%7DL%20%7D%3D%20592.3K)
= 
1 is c 2 is b 3 is 1 d 4 is a
Explanation:
Fe2o3+ 3co ---------. 2fe + 3co2
Answer : The volume of the balloon will be, 0.494 liters
Solution :
Charles' Law : It is defined as the volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
or,
where,
= initial volume of gas = ?
= final volume of gas = 0.54 L
= initial temperature of gas = 273.15 K
= final temperature of gas = 298.5 K
Now put all the given values in the above equation, we get the initial volume of balloon.
Therefore, the volume of the balloon will be, 0.494 liters
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
