Question

How much work is done (in L*atm) by a gas that expands from 7.6 liters to 24 liters against an external pressure of 3.4 atm?

Answer 1

Answer:

The work done by the gas is 55.76 L*atm.

Explanation:

The work can be found as follows:

$W = p\Delta V = p(V_{f} - V_{i})$     (1)

Where:

p: is the pressure = 3.4 atm

V(f): is the final volume = 24 L

V(i): is the initial volume = 7.6 L

By replacing the above values into equation (1) we have:

$W = p(V_{f} - V_{i}) = 3.4 atm(24 L - 7.6 L) = 55.76 atm*L$  

Therefore, the work done by the gas is 55.76 L*atm.

I hope it helps you!