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Chapter 9: Problem 1243

To double the volume of a given mass at an ideal gas at $27^{\circ} \mathrm{C}$ keeping the pressure constant one must raise the temperature in degree centigrade (A) \(54^{\circ}\) (B) \(600^{\circ}\) (C) \(327^{\circ}\) (D) \(270^{\circ}\)

Short Answer

Expert verified
In order to double the volume of a given mass at an ideal gas at \(27^{\circ} \mathrm{C}\) while keeping the pressure constant, one must raise the temperature to \(327^{\circ}\) centigrade.

Step by step solution

01

Write down the Ideal Gas Law formula

The Ideal Gas Law formula is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin. Since the number of moles and the gas constant are the same for both initial and final states, we can simplify this formula by dividing both sides by nR, giving us the equation: PV/T = constant.
02

Understand the information given in the problem

We are given the initial temperature T1 in Celsius and asked to find the final temperature T2 in Celsius while keeping pressure constant and doubling the volume. The initial temperature T1 = 27°C. It's important to convert this temperature to Kelvin scale: \( T1(K) = 27 + 273 = 300 K \).
03

Set up the relationship between the initial and final states

Since we are doubling the volume and keeping the pressure constant, the new volume is 2V. Our equation when considering the initial and final temperatures and volumes becomes: \( \frac{PV}{T_1} = \frac{P(2V)}{T_2} \)
04

Solve for the final temperature in Kelvin

We can now solve the equation for T2: \( \frac{T_1}{2} = \frac{T_2}{T_1} \) \( T_2 = 2 \times T_1 = 2 \times 300 K = 600 K \)
05

Convert the final temperature to Celsius

We need to find the final temperature (T2) in Celsius: \( T_2(°C) = T_2(K) - 273 = 600 - 273 = 327°C \) The final temperature is 327°C, so the correct answer is (C).

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Most popular questions from this chapter

The molar specific heat at constant pressure for a monoatomic gas is (A) \((3 / 2) \mathrm{R}\) (B) \((5 / 2) \mathrm{R}\) (C) \(4 \mathrm{R}\) (D) \((7 / 2) \mathrm{R}\)

The ratio of the vapor densities of two gases at a given temperature is $9: 8$, The ratio of the rms velocities of their molecule is (A) \(3: 2 \sqrt{2}\) (B) \(2 \sqrt{2}: 3\) (C) \(9: 8\) (D) \(8: 9\)

If the ratio of vapor density for hydrogen and oxygen is \([1 /(16)]\), then under constant pressure the ratio of their \(\mathrm{rms}\) velocities will be (A) \(4: 1\) (B) \(1: 16\) (C) \(16: 1\) (D) \(1: 4\)

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

At what temperature is the kinetic energy of a gas molecule double that of its value at \(27^{\circ} \mathrm{C}\) (A) \(54^{\circ} \mathrm{C}\) (B) \(108^{\circ} \mathrm{C}\) (C) \(327^{\circ} \mathrm{C}\) (D) \(300^{\circ} \mathrm{C}\)
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