Equally the same paragraph
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
Answer:
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Explanation:
Because diffusion<span> is the process when molecules move to lower or higher concentration, so as the molecules move faster they are going to lower or high concentration faster.</span>