Question

0.05kg of CO2 with MW=44, occupying a volume of 0.03m³ at 1.025bar, is compressed reversible until the pressure is 6.15bar
Calculate the final temperature, the work done on the CO², the heat flow to or from the cylinder walls
When the process is according to law PV 1.4=constant
When the process is isothermal
When the process takes place in a perfectly thermally insulated cylinder
Assume CO2 to be a perfect gas take 1.3

Answer 1

$p_1V_1=\frac{m}{M}RT_1\to T_1=\frac{p_1V_1M}{m R}=\frac{102500\cdot 0.03\cdot 0.044}{0.05\cdot8.31}\approx326 K$

(a) $pV^{1.4}=const$

$\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{\frac{1.4-1}{1.4}}\to T_2=T_1(\frac{p_2}{p_1})=326\cdot(\frac{615000}{102500})^{\frac{1.4-1}{1.4}}=544K$

$p_1V_1^{1.4}=p_2V_2^{1.4}\to V_2=(\frac{p_1V_1^{1.4}}{p_2})^{1/1.4}=(\frac{102500\cdot 0.03^{1.4}}{615000})^{1/1.4}=0.00834m^3$

$W=\frac{p_1V_1-p_2V_2}{n-1}=\frac{102500\cdot 0.03-615000\cdot 0.00834}{1.4-1}=-5135J$ ; $W_{out}=5135J$

$Q=\Delta U+W=\frac{m}{M}\frac{i}{2}R\Delta T+W=\frac{0.05}{0.044}\cdot\frac{6}{2}\cdot8.31\cdot 218-5135=1041J$

(b) $T=const$

$p_1V_1=p_2V_2\to V_2=\frac{p_1V_1}{p_2}=\frac{102500\cdot 0.03}{615000}=0.005m^3$

$W=\frac{m}{M}RT\ln{\frac{p_1}{p_2}}=\frac{0.05}{0.044}\cdot8.31\cdot 326\cdot ln{\frac{102500}{615000}}=-5516J;$ $W_{out}=5516J$

$Q=\Delta U+W=0-5516=-5516J$

(c) $Q=0$

$\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{\frac{1.3-1}{1.3}}\to T_2=T_1(\frac{p_2}{p_1})=326\cdot(\frac{615000}{102500})^{\frac{1.3-1}{1.3}}=493K$

$p_1V_1^{1.3}=p_2V_2^{1.3}\to V_2=(\frac{p_1V_1^{1.3}}{p_2})^{1/1.3}=(\frac{102500\cdot 0.03^{1.3}}{615000})^{1/1.3}=0.00756m^3$

$W=\frac{p_1V_1-p_2V_2}{n-1}=\frac{102500\cdot 0.03-615000\cdot 0.00756}{1.3-1}=-5248J;$ $W_{out}=5248J$

$Q=0$