Question

Suppose \(15.0 \mathrm{~L}\) of an ideal monatomic gas at a pressure of \(1.50 \cdot 10^{5} \mathrm{kPa}\) is expanded adiabatically (no heat transfer) until the volume is doubled. a) What is the pressure of the gas at the new volume? b) If the initial temperature of the gas was \(300 . \mathrm{K},\) what is its final temperature after the expansion?

Short answer

Based on the given initial conditions and the fact that the ideal monatomic gas undergoes an adiabatic expansion, calculate the final pressure and temperature of the gas. Initial volume (\(V_1\)): 15.0 L Initial pressure (\(P_1\)): 1.50 x 10^5 kPa Initial temperature (\(T_1\)): 300 K Final volume (\(V_2\)): 30.0 L Answer: Final pressure (\(P_2\)): 3.35 x 10^4 kPa Final temperature (\(T_2\)): 134 K

Step-by-step solution

Write down the initial conditions and the adiabatic equation for ideal gases

Given information: Initial volume, \(V_1 = 15.0 L\) Initial pressure, \(P_1 = 1.50 \times 10^5 kPa = 1.50 \times 10^8 Pa\) (1 kPa = 1000 Pa) Initial temperature, \(T_1 = 300 K\) The final volume is twice the initial volume, \(V_2 = 2V_1 = 30.0 L\) To solve the problem, we will use the adiabatic equation for ideal gases: \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \Rightarrow P_2=\frac{P_1 V_1^{\gamma}}{V_2^{\gamma}}\) Where \(P_1\) and \(P_2\) are the initial and final pressures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(\gamma\) is the adiabatic index, which is equal to \(\frac{5}{3}\) for monatomic ideal gases.

Calculate the final pressure using the adiabatic equation

Substituting the given values and \(\gamma=\frac{5}{3}\) into the adiabatic equation: \(P_2 =\frac{P_1 V_1^{\frac{5}{3}}}{V_2^{\frac{5}{3}}}=\frac{(1.50 \times 10^8 Pa)(15.0 L)^{\frac{5}{3}}}{(30.0 L)^{\frac{5}{3}}}\) Solve for \(P_2\): \(P_2 \approx 3.35 \times 10^7 Pa\) or \(3.35 \times 10^4 kPa\) The final pressure of the gas at the new volume is approximately \(3.35 \times 10^4 kPa\).

Use the ideal gas law to find the final temperature

The ideal gas law is given by \(PV=nRT \Rightarrow \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\) We need to find \(T_2\), the final temperature. First, we can solve for \(\frac{P_1 V_1}{T_1}\): \(\frac{P_1 V_1}{T_1}=\frac{(1.50 \times 10^8 Pa)(15.0 L)}{300 K} \approx 7.50 \times 10^6 \frac{Pa L}{K}\) Now, we'll use the final pressure and volume to find \(T_2\): \(T_2=\frac{P_2 V_2}{\frac{P_1 V_1}{T_1}} =\frac{(3.35 \times 10^7 Pa)(30.0 L)}{7.50 \times 10^6 \frac{Pa L}{K}}\) Solve for \(T_2\): \(T_2 \approx 134 K\) The final temperature of the gas after the expansion is approximately \(134 K\).