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.
(A) 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)

Q = 62 MJ

Explanation:

given data

ideal gas mixture He = 3 kmol  

ideal gas mixture Ar = 7 kmol

temperature t1 = 27°C = 300 K

pressure p1 = 200 kPa

solution

we apply here charle's law for pressure p constant that is

v ∝ T   ..................1

and put here value we get

\frac{2v1}{v1} = \frac{t2}{300}  

solve it we get

t2 = 600 K

so here mass of He is

mass = mole of He × molecular weight He

mass = 3 × 4

mass = 12 u

and

mass of Ar is

mass = mole of Ar × molecular weight Ar

mass = 7  × 40  

mass = 280

and

heat supply will be

heat supply = mHe × CpHe × ΔT + mAr × CpAr × ΔT    ................2

and here we use gas table and we get

CpHe = 5.19 (kJ/(kg K)) and CpAr = 0.5203 (kJ/(kg K))

put here value and we get

Q = 12 × 5.19 × 10³ (600-300) + 280 × 0.5203 × 10³ ( 600 -300 )

Q = 62 MJ

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