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Chapter 9: Problem 1248

A gas at \(27{ }^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (A) \(273^{\circ} \mathrm{C}\) (B) \(-173^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature of the gas becomes approximately \(-173^\circ \mathrm{C}\) after the expansion.

Step by step solution

01

Convert all temperatures to Kelvin

The initial temperature is given in Celsius, but we need it in Kelvin for our calculations. To convert Celsius to Kelvin, simply add 273.15. Initial temperature in Celsius: \(T_1 = 27^\circ C\) Convert to Kelvin: \(T_1 = 27 + 273.15 = 300.15 K\)
02

Write down the initial and final pressures and volumes of the gas

The initial pressure is given as 30 atmospheric pressure, and the gas is allowed to expand until it reaches atmospheric pressure (1 atm). Initial pressure: \(P_1 = 30\;\text{atm}\) Final pressure: \(P_2 = 1\;\text{atm}\) The final volume is twice the initial volume, so we can write: Final volume: \(V_2 = 2V_1\)
03

Apply the Combined Gas Law

The Combined Gas Law is given by: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We know all the variables except for the final temperature (\(T_2\)). We can solve this equation for \(T_2\): \(T_2 = \frac{P_2V_2T_1}{P_1V_1}\)
04

Substitute the values and solve for the final temperature

Now we can plug in the known values and solve for \(T_2\): \(T_2 = \frac{(1\;\text{atm})(2V_1)(300.15\;\text{K})}{(30\;\text{atm})(V_1)}\) The initial volume (\(V_1\)) cancels out: \(T_2 = \frac{2(300.15\;\text{K})}{30}\) Calculate the final temperature in Kelvin: \(T_2 = 20.01\;\text{K}\)
05

Convert the final temperature to Celsius

Convert the final temperature from Kelvin to Celsius: Final temperature in Celsius: \(T_2 = 20.01 - 273.15 = - 253.14^\circ C \) The closest answer choice to our calculated value is: (B) \(-173^\circ \mathrm{C}\) The final temperature of the gas becomes approximately -173°C after the expansion.

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Most popular questions from this chapter

At \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

The mean kinetic energy of a gas at \(300 \mathrm{~K}\) is \(100 \mathrm{~J}\). mean energy of the gas at \(450 \mathrm{~K}\) is equal to (A) \(100 \mathrm{~J}\) (B) \(150 \mathrm{~J}\) (C) \(3000 \mathrm{~J}\) (D) \(450 \mathrm{~J}\)

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Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)

At \(0 \mathrm{~K}\) which of the following properties of a gas will be zero (A) Kinetic energy (B) Density (C) Potential energy (D) Vibrational energy
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