Question

One cubic meter of an ideal gas at 600 K and 1000 Malek kPa expands to five times its initial volume as follows; a) By a mechanically reversible, isothermal process, b) By a mechanical reversible, adiabatic process, c) By an adiabatically irreversible process in which expansion is against a restraining pressure of 100 kPa For each case calculate the final temperature, pressure, and the work done by the gas. Cp 21J/mol K.

Answer 1

Answer:

Follows are the solution to this question:

Explanation:

Given value:

$v_1=1m^3 \\ t_1 = 600 \ K \\ p_1 = 1000 \ kpa \\v_1 = 5 \ v_1 = 5 \ m^3$

In point a:

Calculating the process of Isothermal, when the temperature is constant:

$\to T_1 = T_2 = 600 \ K \\\\\to \bold{P_1V_1 = P_2V_2} \\\\\to 1000 \ (Kpa) \times 1 \ (m^3) \neq p_2 \times 5 \ (m^3)\\\\\to p_2 = 200 \ kpa\\\\\to w = nRT \ In (\frac{v_2}{v_1}) = p_1v_1 \ ln ( \frac{v_2}{v_1})$

                               $= 1000 \times 1 \times ln (\frac{5}{1}) \\\\ = 1.61 \ KJ$

In point b:

Calculating the adiobatic process:

$\to p_1v_1^\gamma = p_2v_2^\gamma \\\\ \to \gamma = \frac{c_p}{c_v} \\\\\to R= c_p -c_v \\\\ \to c_p= 21 \frac{J}{mol.k}\\\\\to \gamma = \frac{c_p}{c_p-R}$

       $= \frac{21}{21.8}\\\\ = 1.62\\\\= 1000 \times 1^{1.62}\\\\ = p_2 \times 5^{1.62}$

$p_2 = 73.73 \ Kpa$

$\to p_1^{1-\gamma} t_1^{\gamma} = p_2^{1-\gamma } t_2^{\gamma }$

$= 1000^{1-1.62} \times 600^{1.62} = 73.73^{1-1.62} \times t_2^{\gamma}\\\\ \to t_2^{\gamma} = 6288.5\\\\\to t_2= 6258.5 ^\frac{1}{1.62}$

        $= 221.2 \ k$

In point c:

$\to w= \frac{p_2v_2-p_1v_1}{\gamma -1}$

        $= \frac{(73.73 \times 5)- ( 1000\times 1)}{1.620-1}$

        $= \frac{( -631.35 )}{.620}\\\\= -1018.31 \ KJ$

workdone by gas is 1.018 KJ