Answer:
Q = 62 ( since we are instructed not to include the units in the answer)
Explanation:
Given that:
[tex]n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol[/tex]
[tex]T_1 = 27^0 \ C = ( 27+273)K = 300 K[/tex]
[tex]P_1 = 200 \ kPa[/tex]
Q = ???
Now the gas expands at constant pressure until its volume doubles
i.e if [tex]V_1 = x\\V_2 = 2V_1[/tex]
Using Charles Law; since pressure is constant
[tex]V \alpha T[/tex]
[tex]\frac{V_2}{V_1} =\frac{T_2}{T_1}[/tex]
[tex]\frac{2V_1}{V_1} =\frac{T_2}{300}[/tex]
[tex]T_2 = 300*2\\T_2 = 600[/tex]
mass of He =number of moles of He × molecular weight of He
mass of He = 3 kg × 4
mass of He = 12 kg
mass of Ar =number of moles of Ar × molecular weight of Ar
mass of He = 7 kg × 40
mass of He = 280 kg
Now; the amount of Heat Q transferred = [tex]m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T[/tex]
From gas table
[tex]Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K[/tex]
∴ Q = [tex]12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)[/tex]
Q = [tex]62.389 *10^6[/tex]
Q = 62 MJ
Q = 62 ( since we are instructed not to include the units in the answer)
Answer:
Q = 62
Explanation:
The solution is obtained by manipulating the energy balance of the system:
Q = W + ΔU
⇒ Q = P*(V₂ - V₁) + mcv*(T₂ - T₁)
⇒ Q = P*V₁ + m*(((mcv)He /m) + ((mcv)Ar /m)*((P*V₂/(mR) - T₁)
⇒ Q = mRT₁ + ((mcv)He + (mcv)Ar)*(2P*V₁/(mR) - T₁)
⇒ Q = T₁*(Nm*Ru + (mcv)He + (mcv)Ar)
⇒ Q = T₁*(Nm*Ru + (MNcv)He + (MNcv)Ar)
where
T₁ = (27 + 273) K = 300 K
Nm = (7 + 3) kmol = 10 kmol = 10⁴ mol
Ru = 8.314 J*K⁻¹*mol⁻¹
(MNcv)He = (4 g*mol⁻¹)*(3*10³ mol)*(3.1156 kJ*Kg⁻¹*K⁻¹)*(1 kg/10³ g)*(1 MJ/10³ kJ) = 0.0373872 MJ*K⁻¹
(MNcv)Ar = (40 g*mol⁻¹)*(7*10³ mol)*(0.3122 kJ*Kg⁻¹*K⁻¹)*(1 kg/10³ g)*(1 MJ/10³ kJ) = 0.087416 MJ*K⁻¹
Finally, we get
⇒ Q = 300*(10*8.314 + 4*3*3.1156 + 40*7*0.3122)*10⁻³
⇒ Q = 62
