Thermodynamics and Work Calculations

The video tutorial explains the concept of internal energy and its relation to heat and work, focusing on the first law of thermodynamics. It covers sign conventions in physics and chemistry and provides four problem-solving examples. These examples demonstrate how to calculate changes in internal energy when heat is added or removed and when work is done by or on a system, including gas expansion and compression scenarios.

What does the first law of thermodynamics state about energy?

• Energy is only conserved in closed systems.

• Energy is always lost in a system.

• Energy can be transferred or transformed but not created or destroyed.

• Energy can be created or destroyed.

In physics, if heat energy is added to a system, what is the sign of Q?

• Negative

• Zero

• Undefined

• Positive

In the first problem, what is the change in internal energy when 300 J of heat is added and 200 J of work is done by the system?

• 500 J decrease

• 100 J increase

• 100 J decrease

• 500 J increase

In the second problem, what is the change in internal energy when 700 J of heat is removed and 400 J of work is done on the system?

• 300 J increase

• 300 J decrease

• 1100 J decrease

• 1100 J increase

What is the formula to calculate work done by a gas at constant pressure?

• Work = Volume / Pressure

• Work = Pressure x Volume

• Work = Pressure / Volume

• Work = Pressure x Change in Volume

In the third problem, how much work is done by the gas when it expands from 0.02 m³ to 0.05 m³ at 45,000 Pa?

• 2000 J

• 1000 J

• 1500 J

• 1350 J

In the third problem, what is the change in internal energy when 1500 J of heat is added and 1350 J of work is done by the gas?

• 2850 J increase

• 150 J increase

• 2850 J decrease

• 150 J decrease

In the fourth problem, what is the work done on the gas when it is compressed from 25 L to 15 L at 3 atm?

• 1013 J

• -1013 J

• 3039 J

• -3039 J

What is the conversion factor from liters-atm to joules?

• 1 L-atm = 1.013 J

• 1 L-atm = 101.3 J

• 1 L-atm = 0.1013 J

• 1 L-atm = 10.13 J

In the fourth problem, what is the change in internal energy when 2500 J of heat is removed and 3039 J of work is done on the gas?

• 539 J increase

• 5539 J decrease

• 539 J decrease

• 5539 J increase