Question

A mixture of nitrogen and argon gas is compressed from a volume of 99.0L to a volume of 51.0L , while the pressure is held constant at 61.0atm . Calculate the work done on the gas mixture.

Answer 1

The work done on the gas mixture of nitrogen and argon during compression from 99.0 L to 51.0 L at constant pressure is 296,436 J.

Step-by-step explanation:

To calculate the work done on a gas mixture when it is compressed, we can use the formula W = -PΔV, where W is the work done on the gas, P is the constant pressure, and ΔV is the change in volume. Since the pressure is held constant at 61.0 atm and the volume changes from 99.0 L to 51.0 L, we have ΔV = Vfinal – Vinitial = 51.0 L - 99.0 L = -48.0 L. We must convert the volume to liters to match the pressure units (1 L = 0.001 m³), so ΔV is -0.048 m³. The work done on the gas mixture is therefore:

W = -PΔV = -(61.0 atm * –0.048 m³)

To convert to Joules, we use the conversion factor (1 atm = 101.325 kPa = 101325 Pa), so:

W = -(61.0 * 101325 Pa * –0.048 m³) = 296,436 Joules

The negative sign indicates work is done on the gas. Thus, the work done on the gas mixture is 296,436 J.