Question:

A cylinder containing an ideal gas undergoes a thermodynamic process as shown in the graph below. The process is initially isothermal at a temperature of 300 K, then the gas is compressed adiabatically, and finally, the gas is allowed to expand isobarically.

a) Calculate the work done by the gas during the adiabatic compression process. b) Determine the change in internal energy of the gas during the isobaric expansion process.

Answer:

a) To calculate the work done during the adiabatic process, we can use the equation for work done during an adiabatic process:

Where $P_1$ and $V_1$ are the initial pressure and volume, respectively, $P_2$ and $V_2$ are the final pressure and volume, and $n$ is the polytropic index, which is given by the ratio of specific heats $\frac{C_p}{C_v}$ for an ideal gas. Given that the process is adiabatic, we know that $PV^\gamma = constant$, where $\gamma = \frac{C_p}{C_v}$.

First we need to determine the values of $P_1$, $V_1$, $P_2$, and $V_2$ from the graph:

Initial state: [ P_1 = 1.5 \times 10^5 , Pa ] [ V_1 = 2 , m^3 ]

Final state: [ P_2 = 4.5 \times 10^5 , Pa ] [ V_2 = 1 , m^3 ]

Now we can calculate $n$: [ \gamma = \frac{C_p}{C_v} = \frac{7/2R}{5/2R} = \frac{7}{5} ] [ n = \frac{\gamma - 1}{\gamma} = \frac{7/5 - 1}{7/5} = \frac{2}{7} ]

Now we can find the work done: [ W = \frac{P_1V_1 - P_2V_2}{1 - n} = \frac{(1.5 \times 10^5 , Pa)(2 , m^3) - (4.5 \times 10^5 , Pa)(1 , m^3)}{1 - 2/7} ] [ W = \frac{3 \times 10^5 , J - 4.5 \times 10^5 , J}{5/7} = \frac{-1.5 \times 10^5 , J}{5/7} ] [ W = -2.1 \times 10^5 , J ]

b) The change in internal energy during the isobaric expansion process can be calculated using the equation:

Since the process is isobaric, the work done is given by:

From the graph, we can see that the change in volume $\Delta V$ is $2 \, m^3$. The pressure $P$ during the isobaric process is given as $4.5 \times 10^5 \, Pa$.

Now we can calculate the work done: [ W = (4.5 \times 10^5 , Pa)(2 , m^3) = 9 \times 10^5 , J ]

The change in internal energy is then: [ \Delta U = Q - W ]

Since the process is isobaric, the heat added $Q$ is given by: [ Q = nC_p\Delta T ]

Given that the gas expands isobarically, the temperature change $\Delta T$ can be determined from the graph: [ \Delta T = 3 - 2 = 1 , K ]

Substituting the values, we get: [ Q = nC_p\Delta T = (2 , moles)(\frac{7}{2}R)(1 , K) = 7R ]

Now, we can calculate the change in internal energy: [ \Delta U = Q - W = 7R - 9 \times 10^5 , J ]