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Surface Energy

A mixture of ...

Question

A mixture of gas expands from 0.03 m $^{3}$ to 0.06 m $^{3}$ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is:

30 kJ

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54 kJ

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84 kJ

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none of the above

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Solution

The correct option is C 54 kJ
$Δ E = Δ W + Δ Q$
$Δ W = - P Δ V$
$P = 1 M e g a P a s c a l = 10 b a r = 9.87 a t m$
$Δ V = 0.03 m^{3} = 30 L$
$W = - 30 \times 9.87 L a t m = - 30 \times 9.87 \times 101.33 J = - 30 k J$
Substituting in the above formula we get,
$Δ E = - 30 k J + 84 k J = 54 k J$

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A mixture of gas expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is:

The correct option is C 54 kJΔE=ΔW+ΔQΔW=−PΔVP=1Mega Pascal=10 bar=9.87 atmΔV=0.03 m3=30 LW=−30×9.87 Latm=−30×9.87×101.33J=−30kJSubstituting in the above formula we get,ΔE=−30 kJ+84 kJ=54 kJ

A mixture of gas expands from 0.03 m $^{3}$ to 0.06 m $^{3}$ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is: