Three moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 430 to 160 K.
(a) What is the work done by (or done to) the gas? Include the algebraic sign.
J
(b) What is the change in the internal energy of the gas? Include the algebraic sign.
J
Do b) first using E = 3/2 N k T. Answer to a) then follows from the fact that the internal energy change is 100% due to work performed.
To find the work done by (or done to) the gas, we can use the formula for adiabatic work:
\(W = \frac{(nR)}{(γ - 1)}(T_2 - T_1)\)
Where:
- \(W\) is the work done by (or done to) the gas,
- \(n\) is the number of moles of the gas,
- \(R\) is the ideal gas constant (\(8.314 \, \text{J/(mol K)}\)),
- \(γ\) is the heat capacity ratio (also known as the adiabatic index) for a monatomic ideal gas, which is equal to \(5/3\) for this case,
- \(T_2\) is the final temperature, and
- \(T_1\) is the initial temperature.
(a) Plugging in the values into the formula, we have:
\(W = \frac{(3 \times 8.314)}{(5/3 - 1)}(160 - 430)\)
Simplifying the equation:
\(W = \frac{24.942}{2/3}(-270)\)
\(W = 24.942 \times -270 \times \frac{3}{2}\)
\(W = -5992.98 \, \text{J}\)
Therefore, the work done by (or done to) the gas is \(-5992.98 \, \text{J}\). The negative sign indicates work done on the gas.
To find the change in the internal energy of the gas, we can use the first law of thermodynamics:
\(\Delta U = Q - W\)
Since the process is adiabatic (no heat exchange with the surroundings), \(Q = 0\). Therefore, the equation becomes:
\(\Delta U = -W\)
(b) Plugging in the value of \(W\) we found earlier:
\(\Delta U = -(-5992.98)\)
Simplifying the equation:
\(\Delta U = 5992.98 \, \text{J}\)
Therefore, the change in the internal energy of the gas is \(5992.98 \, \text{J}\). The positive sign indicates an increase in internal energy.