Knowee
Questions
Features
Study Tools

The volume of a system is changed from 17.0 L to 32.0 L under a pressure of 1.23 atm. Calculate the work of the system?

Question

The volume of a system is changed from 17.0 L to 32.0 L under a pressure of 1.23 atm. Calculate the work of the system?

🧐 Not the exact question you are looking for?Go ask a question

Solution

To calculate the work done by the system, we can use the formula for work done during an isobaric process (a process at constant pressure):

W = PΔV

where: W is the work done, P is the pressure, and ΔV is the change in volume.

First, we need to calculate the change in volume (ΔV). This is the final volume (Vf) minus the initial volume (Vi):

ΔV = Vf - Vi ΔV = 32.0 L - 17.0 L ΔV = 15.0 L

Next, we need to convert the pressure from atm to the SI unit of pressure, which is pascals (Pa). 1 atm is approximately equal to 101325 Pa:

P = 1.23 atm * 101325 Pa/atm P = 124,749.75 Pa

Now we can calculate the work done:

W = PΔV W = 124,749.75 Pa * 15.0 L

However, we need to convert the volume from liters to cubic meters (m^3) because 1 m^3 = 1000 L:

W = 124,749.75 Pa * 15.0 L * 1 m^3/1000 L W = 124,749.75 Pa * 0.015 m^3 W = 1,871.24625 J

So, the work done by the system is approximately 1,871.25 Joules.

This problem has been solved

Similar Questions

A gas at a volume of 4 liters is at a pressure of 2 atm. The volume is changed to 16 Liters, what must the new pressure be?

What is the volume of 0.789 moles of gas that exerts 1.15 atm of pressure at 376 K? Question 7Select one:a.4250 Lb.0.0294 Lc.36.5 Ld.21.2 L

litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ=1.40, the work done by air is: (31.4=4.6555) [Take air to be an ideal gas]

The work done by a gas when its 0.5 mole at 27°C expands from 5 lit to 25 lit reversibly and isothermally is -x kJ. Find value of x ?

A diver uses an air tank that contains 75.7 % nitrogen. Determine the partial pressure of the nitrogen at 17.1 m.

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.