The first law of thermodynamics
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What is the formula for work done during a constant pressure volume change?
The work done (W) during a constant pressure (p) volume change (ΔV) is given by: W = pΔV. Remember to use consistent units (Pascals for pressure and m³ for volume to get Joules for work).
What is the sign convention for work done 'by' the gas versus work done 'on' the gas?
Work done 'by' the gas during expansion is considered positive. Work done 'on' the gas during compression is considered negative. This affects the sign in the first law of thermodynamics.
State the first law of thermodynamics in terms of internal energy, heating, and work.
The first law of thermodynamics states: ΔU = q + W, where ΔU is the increase in internal energy of the system, q is the energy transferred to the system by heating, and W is the work done on the system.
If a gas expands and does 500J of work while absorbing 300J of heat, what is the change in internal energy?
Since the gas does work, W = -500J (work is done 'by' the gas). q = +300J (heat absorbed 'by' the system). Therefore, ΔU = 300J + (-500J) = -200J. The internal energy decreases.
Explain the physical meaning of internal energy (U) in the context of thermodynamics.
Internal energy (U) represents the total energy of the particles (atoms, molecules) within a system. It includes kinetic energy (due to motion) and potential energy (due to intermolecular forces). ΔU represents the change in this total energy.
A gas is compressed at a constant pressure of 2.0 x 10⁵ Pa from a volume of 0.50 m³ to 0.25 m³. Calculate the work done on the gas.
ΔV = 0.25 m³ - 0.50 m³ = -0.25 m³. W = pΔV = (2.0 x 10⁵ Pa)(-0.25 m³) = -5.0 x 10⁴ J. Since W is negative, this represents work done on the gas (compression).
What does a positive value of 'q' signify in the first law of thermodynamics?
A positive value of 'q' indicates that energy is being transferred *to* the system as heat. This leads to an increase in the internal energy of the system, assuming work remains constant.
A system's internal energy increases by 400 J, and 100 J of work is done on it. How much heat was transferred, and was it added to or removed from the system?
Using ΔU = q + W, we have 400 J = q + 100 J. Solving for q, we get q = 300 J. Since q is positive, 300 J of heat was added to the system.
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