Since the speed of the bag is constant, it’s kinetic energy remains constant. Since it’s height above the ground is constant, it’s gravitational potential energy is also constant. Since it’s total energy is constant, net work = 0.
A. Molar Mass
Explanation:
Gram molecular mass is the mass in grams of one mole of a molecular substance. Gram molecular mass is the same as molar mass.
Most nonmetallic solids have the property of brittleness.
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
Answer:
Follows are the solution to this question:
Explanation:
Given value:
![v_1=1m^3 \\ t_1 = 600 \ K \\ p_1 = 1000 \ kpa \\v_1 = 5 \ v_1 = 5 \ m^3](https://tex.z-dn.net/?f=v_1%3D1m%5E3%20%5C%5C%20t_1%20%3D%20600%20%5C%20K%20%5C%5C%20p_1%20%3D%201000%20%5C%20kpa%20%5C%5Cv_1%20%3D%205%20%5C%20v_1%20%20%3D%205%20%5C%20m%5E3)
In point a:
Calculating the process of Isothermal, when the temperature is constant:
![\to T_1 = T_2 = 600 \ K \\\\\to \bold{P_1V_1 = P_2V_2} \\\\\to 1000 \ (Kpa) \times 1 \ (m^3) \neq p_2 \times 5 \ (m^3)\\\\\to p_2 = 200 \ kpa\\\\\to w = nRT \ In (\frac{v_2}{v_1}) = p_1v_1 \ ln ( \frac{v_2}{v_1})](https://tex.z-dn.net/?f=%5Cto%20%20T_1%20%3D%20T_2%20%3D%20600%20%5C%20K%20%5C%5C%5C%5C%5Cto%20%5Cbold%7BP_1V_1%20%3D%20P_2V_2%7D%20%5C%5C%5C%5C%5Cto%201000%20%5C%20%28Kpa%29%20%20%5Ctimes%201%20%5C%20%28m%5E3%29%20%5Cneq%20p_2%20%5Ctimes%20%205%20%5C%20%28m%5E3%29%5C%5C%5C%5C%5Cto%20p_2%20%3D%20200%20%5C%20kpa%5C%5C%5C%5C%5Cto%20w%20%3D%20nRT%20%5C%20In%20%28%5Cfrac%7Bv_2%7D%7Bv_1%7D%29%20%3D%20%20p_1v_1%20%20%20%5C%20ln%20%28%20%5Cfrac%7Bv_2%7D%7Bv_1%7D%29)
![= 1000 \times 1 \times ln (\frac{5}{1}) \\\\ = 1.61 \ KJ](https://tex.z-dn.net/?f=%3D%201000%20%5Ctimes%20%201%20%5Ctimes%20ln%20%20%28%5Cfrac%7B5%7D%7B1%7D%29%20%5C%5C%5C%5C%20%3D%201.61%20%5C%20KJ)
In point b:
Calculating the adiobatic process:
![\to p_1v_1^\gamma = p_2v_2^\gamma \\\\ \to \gamma = \frac{c_p}{c_v} \\\\\to R= c_p -c_v \\\\ \to c_p= 21 \frac{J}{mol.k}\\\\\to \gamma = \frac{c_p}{c_p-R} \\\\](https://tex.z-dn.net/?f=%5Cto%20%20p_1v_1%5E%5Cgamma%20%20%3D%20p_2v_2%5E%5Cgamma%20%5C%5C%5C%5C%20%5Cto%20%5Cgamma%20%20%3D%20%5Cfrac%7Bc_p%7D%7Bc_v%7D%20%5C%5C%5C%5C%5Cto%20%20R%3D%20c_p%20-c_v%20%20%5C%5C%5C%5C%20%5Cto%20c_p%3D%2021%20%5Cfrac%7BJ%7D%7Bmol.k%7D%5C%5C%5C%5C%5Cto%20%5Cgamma%20%20%3D%20%5Cfrac%7Bc_p%7D%7Bc_p-R%7D%20%5C%5C%5C%5C)
![= \frac{21}{21.8}\\\\ = 1.62\\\\= 1000 \times 1^{1.62}\\\\ = p_2 \times 5^{1.62}\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B21%7D%7B21.8%7D%5C%5C%5C%5C%20%3D%201.62%5C%5C%5C%5C%3D%201000%20%5Ctimes%201%5E%7B1.62%7D%5C%5C%5C%5C%20%3D%20p_2%20%5Ctimes%20%205%5E%7B1.62%7D%5C%5C%5C%5C)
![p_2 = 73.73 \ Kpa](https://tex.z-dn.net/?f=p_2%20%3D%20%2073.73%20%5C%20Kpa)
![\to p_1^{1-\gamma} t_1^{\gamma} = p_2^{1-\gamma } t_2^{\gamma }](https://tex.z-dn.net/?f=%5Cto%20p_1%5E%7B1-%5Cgamma%7D%20t_1%5E%7B%5Cgamma%7D%20%3D%20p_2%5E%7B1-%5Cgamma%20%7D%20t_2%5E%7B%5Cgamma%20%7D)
![= 1000^{1-1.62} \times 600^{1.62} = 73.73^{1-1.62} \times t_2^{\gamma}\\\\ \to t_2^{\gamma} = 6288.5\\\\\to t_2= 6258.5 ^\frac{1}{1.62} \\\\](https://tex.z-dn.net/?f=%3D%201000%5E%7B1-1.62%7D%20%5Ctimes%20600%5E%7B1.62%7D%20%3D%2073.73%5E%7B1-1.62%7D%20%5Ctimes%20t_2%5E%7B%5Cgamma%7D%5C%5C%5C%5C%20%5Cto%20t_2%5E%7B%5Cgamma%7D%20%3D%206288.5%5C%5C%5C%5C%5Cto%20t_2%3D%206258.5%20%5E%5Cfrac%7B1%7D%7B1.62%7D%20%5C%5C%5C%5C)
![= 221.2 \ k](https://tex.z-dn.net/?f=%3D%20221.2%20%5C%20k)
In point c:
![\to w= \frac{p_2v_2-p_1v_1}{\gamma -1}](https://tex.z-dn.net/?f=%5Cto%20w%3D%20%5Cfrac%7Bp_2v_2-p_1v_1%7D%7B%5Cgamma%20-1%7D)
![= \frac{(73.73 \times 5)- ( 1000\times 1)}{1.620-1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%2873.73%20%5Ctimes%205%29-%20%28%201000%5Ctimes%201%29%7D%7B1.620-1%7D)
![= \frac{( -631.35 )}{.620}\\\\= -1018.31 \ KJ](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%20-631.35%20%29%7D%7B.620%7D%5C%5C%5C%5C%3D%20-1018.31%20%5C%20KJ)
workdone by gas is 1.018 KJ