Answer to Question #194071 in Mechanical Engineering for Brian

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 final temperature, the work done on the CO²,the heat flow to or form the cylinder walls

(i)when the process is according to law pV¹.⁴=constant

(ii)when the process is Isothermal

(iii)when the process takes place in a perfectly thermal insulated cylinder.

Assume CO² to be a perfect gas takes Y=1.3

Answer 1

The mass is $m = 0.05\,\mathrm{kg},$ $\mu=44\,\mathrm{g/mol}$ , the initial volume is $V_1=0.03\,\mathrm{m}^3,$ the initial pressure is $p_1 = 1.025\,\mathrm{bar} = 1.025\cdot10^5\,\mathrm{Pa}$ . The final pressure is $p_2= 6.15\,\mathrm{bar}=6.15\cdot10^5\,\mathrm{Pa}.$

(i) If $pV^{1.4}=\text{constant},$ then $p_1V_1^{1.4} = p_2V_2^{1.4} \; \Rightarrow \; V_2 = V_1\left( \dfrac{p_1}{p_2}\right)^{1/1.4} = 8.3\cdot10^{-3}\,\mathrm{m^3}.$ For such a polytropic process the work can be calculated as

$A = \int p\,dV =\nu \dfrac{R}{1.4-1}\cdot T_1\left( 1-\dfrac{T_2}{T_1}\right)$ .

$p_1V_1 = \dfrac{m}{\mu}RT_1, \; T_1 = 325.6\,\mathrm{K} ; \;\; p_2V_2 = \dfrac{m}{\mu}RT_2, \; T_2 = 540.5\,\mathrm{K}.$

Therefore, the work of gas will be $A =\nu \dfrac{R}{1.4-1}\cdot T_1\left( 1-\dfrac{T_2}{T_1}\right) = -5073\,\mathrm{J}.$

The heat that should be given to gas can be calculated as

$q=\int T\,ds =\nu c_v \dfrac{1.4-1.3}{1.4-1}(T_2-T_1) = 3R\cdot \nu\cdot \dfrac{1.4-1.3}{1.4-1}(540.5-325.6) = 1522\,\mathrm{J}.$

(ii) In the isothermal process $T_1=T_2 = 325.6\,\mathrm{K}.$ For such process work can be calculated as

$A = \int p\,dV =\nu R T\ln \dfrac{V_2}{V_1} = \nu R T\ln \dfrac{p_1}{p_2} = -5509\,\mathrm{J}.$ The change of internal energy is 0, so the heat that should be given to the gas is $-5509\mathrm{J}.$

(iii) if the cylinder is thermally insulated the heat flow will be 0, so the process will be adiabatic. Let us calculate the final temperature:

$pV^{1.3}=\text{constant}, \;\; p_1V_1^{1.3} = p_2V_2^{1.3} \; \Rightarrow \; V_2 = V_1\left( \dfrac{p_1}{p_2}\right)^{1/1.3} = 7.56\cdot10^{-3}\,\mathrm{m^3}.$

$p_2V_2 = \dfrac{m}{\mu}RT_2, \; T_2 = 492.4\,\mathrm{K}.$

The work of the gas is equal to $A = - (U_2-U_1) = c_v\nu (T_1-T_2) = 3R\nu (T_1-T_2) = -4725\,\mathrm{J}.$

We may notice that the work of the gas is negative, so we should do work on gas