5.4: First Law of Thermodynamics
- Page ID
- 167654
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Internal Energy
Exercise \(\PageIndex{1}\)
Calculate ΔU of a gas for a process in which the gas absorbs 60 J of heat and does 15 J of work by expanding.
- Answer
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Δ U = 45 J
\(q = + 60 J\) (positive, system absorbs heat)
\(w = - 15 J\) (negative, system does work)
\(\Delta U = q + w = 60 J - 15 J = 45 J\)
Exercise \(\PageIndex{2}\)
What is the change in internal energy of the system (ΔU) if 110 kJ of heat energy is evolved by the system and 62 kJ of work is done on the system for a certain process?
- Answer
-
Δ U = -48 kJ
\(q = - 110 J\) (negative, system releases heat)
\(w = + 62 J\) (positive, system has work done on it)
\(\Delta U = q + w = -110 J + 62 J = - 48 J\)
Pressure-Volume Work and Enthalpy
Exercise \(\PageIndex{3}\)
A sample of an ideal gas in the cylinder of an engine is compressed from 750.0 mL to 30.5 mL during the compression stroke against a constant pressure of 9.05 x 105 Pa. At the same time, 250 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy ΔU of the gas in joules?
- Answer
-
Δ U = 401 J
\(\Delta U = \Delta H -P\Delta V\)
\(\Delta H = -250 J\)
\(W = P\Delta V\)
\(\Delta V = \left ( 0.0305 L - 0.750 L \right ) = -0.7195 L * \left ( \frac{1 m^{3}}{1000 L} \right ) = -7.195*10^{-4} m^{3}\)
Note: A Pascal is also defined as kg/m*s2. A joule is defined as kg*m2/s2
\(W = \left ( 9.05 * 10^{5} Kg/m*s^{-2} \right )\left ( -7.195*10^{-4} m^{3} \right ) = -651.1475 J\)
\(\Delta U = \Delta H + P\Delta V\)
\[\Delta U = -250 J - (-651 J) = 401 J\]