Question

e 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.

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