Respuesta :
Answer:
Follows are the solution to this question:
Explanation:
Given value:
[tex]v_1=1m^3 \\ t_1 = 600 \ K \\ p_1 = 1000 \ kpa \\v_1 = 5 \ v_1 = 5 \ m^3[/tex]
In point a:
Calculating the process of Isothermal, when the temperature is constant:
[tex]\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})[/tex]
[tex]= 1000 \times 1 \times ln (\frac{5}{1}) \\\\ = 1.61 \ KJ[/tex]
In point b:
Calculating the adiobatic process:
[tex]\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} \\\\[/tex]
[tex]= \frac{21}{21.8}\\\\ = 1.62\\\\= 1000 \times 1^{1.62}\\\\ = p_2 \times 5^{1.62}\\\\[/tex]
[tex]p_2 = 73.73 \ Kpa[/tex]
[tex]\to p_1^{1-\gamma} t_1^{\gamma} = p_2^{1-\gamma } t_2^{\gamma }[/tex]
[tex]= 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} \\\\[/tex]
[tex]= 221.2 \ k[/tex]
In point c:
[tex]\to w= \frac{p_2v_2-p_1v_1}{\gamma -1}[/tex]
[tex]= \frac{(73.73 \times 5)- ( 1000\times 1)}{1.620-1}[/tex]
[tex]= \frac{( -631.35 )}{.620}\\\\= -1018.31 \ KJ[/tex]
workdone by gas is 1.018 KJ
