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Chapter 23: Problem 29

Gas occupies a volume of \(4.33 \mathrm{~L}\) at a pressure of \(1.17 \mathrm{~atm}\) and a temperature of \(310 \mathrm{~K}\). It is compressed adiabatically to a volume of \(1.06 \mathrm{~L}\). Determine \((a)\) the final pressure and \((b)\) the final temperature, assuming the gas to be an ideal gas for which \(\gamma=1.40 .(c)\) How much work was done on the gas?

Short Answer

Expert verified
The final pressure, final temperature, and work done on the gas will be calculated based on the given initial conditions and the formulas applicable to an adiabatic process.

Step by step solution

01

Determine the final pressure according to the adiabatic process formula

We are given the initial pressure \(P_1 = 1.17 \) atm, the initial volume \(V_1 = 4.33 \) L, the final volume \(V_2 = 1.06 \) L, and the specific heat ratio \(\gamma = 1.4\). We can substitute those values into the adiabatic process formula \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\) and solve it for the final pressure \(P_2\). This gives \(P_2 = P_1V_1^{\gamma} / V_2^{\gamma}\).
02

Determine the final temperature of the gas.

The ideal gas law equation can assist in obtaining the final temperature of the gas. The equation is \(P_1V_1 /T_1 = P_2V_2 / T_2\) where \(P\) is the pressure, \(V\) is the volume and \(T\) is the temperature. We know the values for \(P_1\), \(V_1\), \(T_1\), \(P_2\), and \(V_2\). Therefore, we can solve the equation for the final temperature, \(T_2\). This gives \(T_2 = P_2V_2T_1 / (P_1V_1)\).
03

Calculate the work done on the gas.

The work done (\(W\)) can be computed using the formula \(W = (P_2V_2 - P_1V_1) / (1 - \gamma)\). Here, we can input the final pressure, \(P_2\) and volume, \(V_2\), that we computed in Steps 1 and 2, along with the initial pressure, \(P_1\) and volume, \(V_1\), and the specific heat ratio, \(\gamma\), which was given, and compute the amount of work done on the gas.

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