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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

Air is filled in a bottle at atmospheric pressure and it is corked at \(35^{\circ} \mathrm{C}\), If the cork can come out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to remove the cork. (A) \(325.5^{\circ} \mathrm{C}\) (B) \(651^{\circ} \mathrm{C}\) (C) \(851^{\circ} \mathrm{C}\) (D) None of these

What is the mass of 2 liters of nitrogen at \(22.4\) atmospheric pressure and \(273 \mathrm{~K} .\left(\mathrm{R}=8.314 \mathrm{Jmol} \mathrm{k}^{-1}\right)\) (A) \(14 \times 22.4 \mathrm{~g}\) (B) \(56 \mathrm{~g}\) (C) \(28 \mathrm{~g}\) (D) none of these

Hydrogen gas is filled in a balloon at \(20^{\circ} \mathrm{C}\). If temperature is made \(40^{\circ} \mathrm{C}\), pressure remaining the same what fraction of hydrogen will come out (A) \(0.75\) (B) \(0.07\) (C) \(0.25\) D) \(0.5\)

The pressure and temperature of an ideal gas in a closed vessel are $720 \mathrm{kPa}\( and \)40^{\circ} \mathrm{C}\( respectively. If \)(1 / 4)^{\text {th }}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to \(353^{\circ} \mathrm{C}\), final pressure of the gas is (A) \(1440 \mathrm{kPa}\) (B) \(540 \mathrm{kPa}\) (C) \(1080 \mathrm{kPa}\) (D) \(720 \mathrm{kPa}\)

What is the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm and temperature $17^{\circ} \mathrm{C} ?$ Take the radius of nitrogen molecule to be 1A. Molecular mass of nitrogen \(=28\), $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, 1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$ (A) \(2.2 \times 10^{-7} \mathrm{~m}, 2.58 \times 10^{9}\) (B) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{8}\) (C) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{9}\) (D) \(2.2 \times 10^{-7} \mathrm{~m}, 3.58 \times 10^{9}\)
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