Question

Calculate the work in kilojoules done during a synthesis of ammonia in which the volume contracts from 7.4 l to 4.5 l at a constant external pressure of 44 atm . express your answer using two significant figures.

Answer 1

The work done by the gas can be derived from this formula: Work done = P dV ( this is the same as PdeltaV) Now remember P = 44atm and Now dV ( deltaV) = 4.5 - 7.4 = -2.9L ( its contracting) = -2.9 x 10 ^-3m^3 Now remember 1 atm = 101300 pascals Therefore work done by gas = P dV = 44 * -2.9 * x 10 ^-3 *101300 Therefore the above result becomes: = -12925.88 = -13000J to 2 significant figures

Answer 2

13 kilojoules

Further explanation

Given:

A synthesis of ammonia in which the volume contracts from 7.4 l to 4.5 l at a constant external pressure of 44 atm.

Question:

Calculate the work in kilojoules done during the synthesis of ammonia. Express your answer using two significant figures.

The Process:

Let us calculate the change in volume and also convert units.

$\boxed{ \ \Delta V = V_2 - V_1 \ }$

$\boxed{ \ \Delta V = 4.5 - 7.4 \ }$

Therefore ΔV = - 2.9 L.

• Since 1 L = 10⁻³ m³, then ΔV = - 2.9 x 10⁻³ m³.
• Since 1 atm = 1.013 x 10⁵ Pa, then external pressure is p = 44 x 1.013 x 10⁵ Pa, i.e., p = 4,457,200 Pa.  

Work done during expansion and contraction at constant pressure.

$\boxed{ \ W = p \ \Delta V \ }$

Let us find out the work done.

$\boxed{ \ W = 4,457,200 \times -2.9 \times 10^{-3} \ }$

W = -12,925.88 joules

W = -13,000 joules to 2 significant figure.

Thus, the work in kilojoules done during the synthesis of ammonia is 13 kilojoules.

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Notes

• If a gas increases its volume (in other words, ΔV is positive), then the gas does work (ΔW is positive). When a gas expands, it does work on the surroundings.
• In our case, when a gas is compressed, work is done on the gas (in other words, ΔV is negative). The surroundings do work on it. Work is done on the gas (ΔW is negative).

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Notes:

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