How much work, in joules, is done when the pressure is suddenly released from 2.05 atm to 1.87 atm on 27.5 g of He at 27°C?

Question

Quizplus · Accepted Answer

The work done by the system can be calculated using the equation, W = -PΔV, where P is the pressure, ΔV is the change in volume, and the negative sign indicates that work is done on the system.

First, we need to calculate the initial and final volumes of the gas. We can use the ideal gas law to do this:

PV = nRT

where V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Rearranging this equation, we get:

V = (nRT)/P

We can assume that the number of moles of He remains constant throughout the process, so we can use the initial pressure and temperature to calculate the initial volume:

V1 = (nRT1)/P1 = (0.0275 kg)/(4.008 kg/mol x 2.05 atm) x (0.08206 L x atm/K x mol) x (300 K) = 1.37 L

Similarly, we can use the final pressure and temperature to calculate the final volume:

V2 = (nRT2)/P2 = (0.0275 kg)/(4.008 kg/mol x 1.87 atm) x (0.08206 L x atm/K x mol) x (300 K) = 1.50 L

The change in volume, ΔV, is therefore:

ΔV = V2 - V1 = 0.13 L

Now we can use the pressure and volume values to calculate the work done by the system:

W = -PΔV = -(2.05 atm - 1.87 atm) x (101.3 J/L x atm) x (0.13 L) = -1.5 x 10^3 J

Note that the negative sign indicates that work is done on the system, which makes sense since the pressure is suddenly released and the volume increases. Therefore, the best choice is (D) -1.5 × 10^3 J.

How Much Work, in Joules, Is Done When the Pressure