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3 years ago

15

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

°C and 200 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is, in MJ (round to nearest integer; for example if the answer is 53.7MJ, write 54; if the answer is 52.1MJ write 52; do not include the units in your answer)

2 answers:

lidiya [134]3 years ago

5 0

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

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 = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

vfiekz [6]3 years ago

4 0

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

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$h_{o}$ = specific enthalpy at outlet

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