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Chapter 12: Problem 81

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume \(V_{1}\) and contains ideal gas at pressure \(P_{1}\) and temperature \(T_{1}\). The other chamber has volume \(V_{2}\) and contains ideal gas at pressure \(P_{2}\) and temperature \(T_{2}\). If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be: (A) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}\) (B) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}\) (C) \(\frac{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}{P_{1} V_{1}+P_{2} V_{2}}\) (D) \(\frac{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}{P_{1} V_{1}+P_{2} V_{2}}\)

Short Answer

Expert verified
The final equilibrium temperature of the gas in the container is given by: \(T_f = \frac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}\) (Option B).

Step by step solution

01

Recall the ideal gas law

The ideal gas law can be represented as: \(PV = nRT\) Where P is the pressure of the gas, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature of the gas.
02

Find the number of moles for each chamber

We can use the ideal gas law to find the number of moles of gas in each chamber: For chamber 1: \(n_1 = \frac{P_1V_1}{RT_1}\) For chamber 2: \(n_2 = \frac{P_2V_2}{RT_2}\)
03

Apply the conservation of energy

The internal energy of an ideal gas depends only on its temperature. Since the container is insulated, the total energy remains constant during the process. So, the total internal energy before the partition removal is equal to the total internal energy after the removal: \(n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_f\) Where \(C_v\) is the constant volume specific heat capacity of the gas, and \(T_f\) is the final equilibrium temperature.
04

Solve for the final equilibrium temperature

Notice that \(C_v\) can be canceled out on both sides of the equation. Then, substitute the expressions for \(n_1\) and \(n_2\) from Step 2 and solve for \(T_f\): \(\frac{P_1V_1}{RT_1} T_1 + \frac{P_2V_2}{RT_2} T_2 = \left(\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}\right)T_f\) Now, multiply both sides of the equation by \(\frac{RT_1T_2}{P_1V_1+P_2V_2}\): \((T_1T_2)(P_1V_1+P_2V_2) = (P_1V_1T_1+P_2V_2T_2)T_f\) Finally, solve for \(T_f\): \(T_f = \frac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}\) Comparing with the given options, the final equilibrium temperature of the gas in the container is given by option (B).

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Most popular questions from this chapter

Three perfect gases at absolute temperatures, \(T_{1}, T_{2}\), and \(T_{3}\), are mixed. The masses of molecules are \(m_{1}, m_{2}\), and \(m_{3}\) and the number of molecules are \(n_{1}, n_{2}\), and \(n_{3}\), respectively. Assuming no loss of energy, the final temperature of the mixture is [2011] (A) \(\frac{\left(T_{1}+T_{2}+T_{3}\right)}{3}\) (B) \(\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}\) (C) \(\frac{n_{1} T_{1}^{2}+n_{2} T_{2}^{2}+n_{3} T_{3}^{3}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}\) (D) \(\frac{n_{1}^{2} T_{1}^{2}+n_{2}^{2} T_{2}^{2}+n_{3}^{3} T_{3}^{3}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}\)

If amount of heat given to a system is \(50 \mathrm{~J}\) and work done on the system is \(15 \mathrm{~J}\), then change in internal energy of the system is (A) \(35 \mathrm{~J}\) (B) \(50 \mathrm{~J}\) (C) \(65 \mathrm{~J}\) (D) \(15 \mathrm{~J}\)

A Carnot engine, having an efficiency of \(\eta=\frac{1}{10}\) as heat engine is used as a refrigerator. If the work done on the system is \(10 \mathrm{~J}\), the amount of energy absorbed from the reservoir at lower temperature is [2007] (A) \(99 \mathrm{~J}\) (B) \(90 \mathrm{~J}\) (C) \(1 \mathrm{~J}\) (D) \(100 \mathrm{~J}\)

A thermally insulated chamber of volume \(2 V_{0}\) is divided by a frictionless piston of area \(S\) into two equal parts, \(A\) and \(B\). Part \(A\) has an ideal gas at pressure \(P_{0}\) and temperature \(T_{0}\), and in part \(B\) is vacuum. A massless spring of force constant \(k\) is connected with piston and the wall of the container is as shown. Initially, spring is unstretched. Gas in chamber \(A\) is allowed to expand. Let the equilibrium spring be compressed by \(x_{0}\). Then (A) Final pressure of the gas is \(\frac{k x_{0}}{S}\). (B) Work done by the gas is \(\frac{1}{2} k x_{0}^{2}\). (C) Change in internal energy of the gas is \(\frac{1}{2} k x_{0}^{2}\). (D) Temperature of the gas is decreased.

When a system is taken from state \(i\) to state \(f\) along the path iaf, it is found \(Q=50 \mathrm{cal}\) and \(W=20\) cal. Along the path \(i b f, Q=36\) cal. \(W\) along the path \(i b f\) is [2007] (A) \(6 \mathrm{cal}\) (B) \(16 \mathrm{cal}\) (C) \(66 \mathrm{cal}\) (D) \(14 \mathrm{cal}\)
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