Question

If a gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of 2 X 105 N m-2, then change in internal energy is

Answer 1

Answer : The change in internal energy is 100 J.

Solution : Given,

Heat absorbed = 200 J

Change in volume = $500Cm^3$

Pressure = $2\times 10^5Nm^{-2}$

Formula used for change in internal energy when gas expands:

$\Delta U=q-w\\\Delta U=q-P\Delta V$

Now put all the given values in this formula, we get

$\Delta U=200J-[(2\times 10^5Nm^{-2})\times (500Cm^3)]$

$\Delta U=200J-[(2\times 10^5Nm^{-2})\times (5\times 10^{-4}m^3)]=200J-100Nm=200J-100J=100J$

(Conversion : $1Cm^3=10^{-6}m^3$ and $1Nm=1J$)

Therefore, the change in internal energy is 100 J.

Answer 2

Answer:

Explanation:

q=200 J

ΔV=500 cm3

p=2×105Nm−2

1 cm=10−2m

ΔV=500×10−6m3

w=−PΔV

=−2×105N/m2×500×10−6m3

=−100 Nm

w=−100 J

ΔE=+100 J