Q. 100 ml of a liquid is contained in an insulated container at pressure 1 bar.The pressure is steeply increased to 100 bar.Volume of the liquid is decreased by 1 ml at this pressure.Find dH and dU.

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Since the container is insulated, the changes are happening adiabatically - that is, there is no exchange of energy and matter with the environment.

The change can be visualized as having two steps, happening adiabatically. The first step is an increase in pressure (from 1 bar to 100 bar). The second step is compression (from 100 mL to 99 mL, a decrease by 1 mL).

The formulas we need are the following:

`dU = q + w` ; q is heat, while w is work

`dH = dU + pdV + dpV` ; p is pressure, while V is volume

The first step doesn't involve expansion work, but is simply an increase in pressure. Moreover, no heat is added to the system (isolated). Hence, `dU_1 = q + w = 0 + 0 = 0`. For the second step, it's still adiabatic so no heat is added or espaced from the system, but expansion is now present. Expansion can be calculated as follows: `w = -pDeltaV = -100b a r(99mL-100mL) = 100 b a r cdot mL.```

Hence, `dU_2 = q + w = 0 + 100mL = 0.1 b a r *L`

For the entire process, then, `dU = 0.1 b a r * L.`

We use these values to calculate for `dH` . For the first step:

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`dH_1 = dU_1 + pdV + dpV = 0 + 0 + 99 b a r times 100mL`

Hence, `dH_1 = 9900 b a r * mL = 9.9 b a r * L` .

For the second step, and using the same formula, we get:

`dH_2 = 100 b a r * mL + 100 b a r * (-1 mL) + 0 = 0`

Hence, `dH = dH_1 + dH_2 = 9.9 b a r * L + 0 = 9.9 b a r*L`

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Note that 1 bar*L = 1 joule.

Hence, in joules, we have:

dH = 9.9 joules and dU = 0.1 joules.

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