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Chapter 12: Problem 45

Consider two containers \(A\) and \(B\) containing identical gases at the same pressure, volume, and temperature. The gas in container \(A\) is compressed to half of its original volume isothermally while the gas in container \(B\) is compressed to half of its original value adiabatically. The ratio of final pressure of gas in \(B\) to that of gas in \(A\) is (A) \(2^{\gamma-1}\) (B) \(\left(\frac{1}{2}\right)^{\gamma-1}\) (C) \(\left(\frac{1}{1-\gamma}\right)^{2}\) (D) \(\left(\frac{1}{\gamma-1}\right)^{2}\)

Short Answer

Expert verified
The short answer is: (A) \(2^{\gamma-1}\).

Step by step solution

01

Find the final pressure in container A (Isothermal compression)

For an isothermal process happening to an ideal gas, the equation is: \(PV = constant\) Since the volume is compressed to half of its original value and the process is isothermal, we can write: \(P_A * V_A = P_{A_{f}} * \frac{V_A}{2}\) Now, solve for the final pressure, \(P_{A_{f}}\): \(P_{A_{f}} = 2P_A\)
02

Find the final pressure in container B (Adiabatic compression)

For an adiabatic process happening to an ideal gas, the equation is: \(PV^\gamma = constant\) Where \(纬\) is the adiabatic index, \(纬 = \frac{C_P}{C_V}\), and is greater than 1. Since the volume is compressed to half of its original value and the process is adiabatic, we can write: \(P_B * V_B^\gamma = P_{B_{f}} * \left(\frac{V_B}{2}\right)^\gamma\) Now, solve for the final pressure, \(P_{B_{f}}\): \(P_{B_{f}} = P_B * 2^\gamma\)
03

Calculate the ratio of final pressures

To find the ratio of the final pressure of the gas in container B to that of gas in container A, we can write: \(R = \frac{P_{B_{f}}}{P_{A_{f}}}\) Substitute the expressions for final pressures, \(P_{A_{f}}\) and \(P_{B_{f}}\), we found earlier: \(R = \frac{P_B * 2^\gamma}{2P_A}\) Since the initial conditions of the gases in both containers are identical, we can write, \(P_A = P_B\), and therefore: \(R = 2^{\gamma-1}\) So the correct answer is (A) \(2^{\gamma-1}\).

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Most popular questions from this chapter

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio \(C_{P} / C_{V}\) for the gas is [2003] (A) \(4 / 3\) (B) 2 (C) \(5 / 3\) (D) \(3 / 2\)

The work of \(146 \mathrm{~kJ}\) is performed in order to compress 1 kilo mole of a gas adiabatically, and in this process, the temperature of the gas increases by \(7^{\circ} \mathrm{C}\). The gas is \(\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (A) diatomic. (B) triatomic. (C) a mixture of monoatomic and diatomic. (D) monatomic.

One mole of an ideal gas is enclosed in a cylinder fitted with a frictionless piston and occupies a volume of \(1.5\) litre at a pressure of \(1.2 \mathrm{~atm} .\) It is subjected to a process given by equation \(T=\alpha V^{2}, \gamma\) (adiabatic constant) for the gas \(=1.5\). Choose the wrong statement. Given \(R \alpha=80 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{lit}^{-2}(R=\) gas constant and \(\alpha\) is constant) (A) The \(P-V\) diagram of the process is a straight line. (B) The work done by the gas in increasing the volume of the gas to 9 litre is \(3150 \mathrm{~J}\). (C) The change in the internal energy of the gas is \(12600 \mathrm{~J}\). (D) The heat supplied to the gas in the process is 1575 J.

When a given quantity of an ideal monoatomic gas is at pressure \(P\) and absolute temperature \(T\), then the adiabatic bulk modulus of the gas will be (A) \(P\) (B) \(\frac{5}{3} P\) (C) \(T\) (D) \(\frac{5 T}{2}\)

The net work done on the gas in the cycle \(A B C D A\) is \([2009]\) (A) Zero (B) \(276 R\) (C) \(1076 R\) (D) \(1904 R\)
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