Question

Calculate the work (kJ) done during a reaction in which the internal volume expands from 28 L to 51 L against an outside pressure of 4.9 atm. Calculate the work (kJ) done during a reaction in which the internal volume expands from 28 L to 51 L against an outside pressure of 4.9 atm. 11 kJ -11 kJ -39 kJ 39 kJ 0 kJ; No work is done.

Answer 1

Answer:

-11 kJ

Explanation:

Step 1: Convert the pressure to Pascal (SI unit)

We will use the relationship 1 atm = 101,325 Pa.

$4.9 atm \times \frac{101,325Pa}{1atm} = 5.0 \times 10^{5} Pa$

Step 2: Convert the volumes to cubic meters (SI unit)

We will use the relationship 1 m³ = 1,000 L.

$28 L \times \frac{1m^{3} }{1,000L} = 0.028 m^{3}$

$51 L \times \frac{1m^{3} }{1,000L} = 0.051 m^{3}$

Step 3: Calculate the work done during the expansion of a gas

We will use the following expression.

$W = -P \times \Delta V = -5.0 \times 10^{5} Pa \times (0.051m^{3} -0.028m^{3}) =-1.1 \times 10^{4} J$

Step 4: Convert the work to kiloJoule

We will use the relationship 1 kJ = 1,000 J.

$-1.1 \times 10^{4} J \times \frac{1kJ}{1,000J} =-11 kJ$

Answer 2

The work done by the gas is -11 kJ .

The work done by a gas is given by the formula;

w = pΔv

p = pressure of the gas

Δv = change in the volume of the gas

From the question;

p = 4.9 atm

Δv = 51 L - 28 L

Δv =  23 L

w = 4.9 atm ×  23 L

w = 112.7 atmL

But;

1 L atm =101.325 J

112.7 atmL = 112.7 atmL ×  101.325 J/1 L atm

= -11 kJ

Note that the work done is negative because the gas expands and does work on the surroundings

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