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Chapter 9: Problem 1236

A cylinder contains \(10 \mathrm{~kg}\) of gas at pressure of $10^{\prime} \mathrm{N} / \mathrm{m}^{2}$. The quantity of gas taken out of the cylinder, if final pressure is \(2.5 \times 10^{6} \mathrm{Nm}^{-2}\). will be (temperature of gas is constant) (A) \(5.2 \mathrm{~kg}\) (B) \(3.7 \mathrm{~kg}\) (C) \(7.5 \mathrm{~kg}\) (D) \(1 \mathrm{~kg}\)

Short Answer

Expert verified
The short answer is: None of the given options are correct. The mass of the gas removed from the cylinder should be \(9.6\, kg\).

Step by step solution

01

Write the given values

We have been given the following information: Initial mass of gas, \(m_i = 10\, kg\) Initial pressure, \(P_i = 10^5\, N/m^2\) Final pressure of gas, \(P_f = 2.5 \times 10^6\, N/m^2\)
02

Apply ideal gas law

Since the temperature is constant, the ratio of the initial and final pressure and volume remains the same. So, we can write the following equation: \(\frac{P_i V_i}{P_f V_f} = \frac{n_i}{n_f}\) Where \(V_i\) and \(V_f\) are the initial and final volume, and \(n_i\) and \(n_f\) are the initial and final number of moles of gas, respectively.
03

Solve for the initial and final volume ratio

Using the equation from step 2, we can solve for the ratio of initial and final volume: \(\frac{V_i}{V_f} = \frac{P_f}{P_i} = \frac{2.5 \times 10^6}{10^5} = 25\)
04

Solve for the mass of gas removed

Since mass and volume are proportional (mass = density * volume), the ratio of the initial and final mass is the same as the ratio of initial and final volume, which we found in step 3: \(\frac{m_i}{m_f} = \frac{V_i}{V_f} = 25\) Now, we can solve for the final mass, \(m_f\): \(m_f = \frac{m_i}{25} = \frac{10\, kg}{25} = 0.4\, kg\) Finally, we can find the mass of the gas removed, \(m_r\), by subtracting the final mass from the initial mass: \(m_r = m_i - m_f = 10\, kg - 0.4\, kg = 9.6\, kg\) However, this result is not available in the given options. It is possible that there was a misprint in the question or the answer choices. Based on the method we have used, the answer should be \(9.6\, kg\).

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Most popular questions from this chapter

If the degrees of freedom of a gas are \(\mathrm{f}\), then the ratio of two specific heats $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)$ is given by (A) \(1-(2 / \mathrm{f})\) (B) \(1+(1 / f)\) (C) \((2 / \mathrm{f})+1\) (D) \(1+(3 / \mathrm{f})\)

Calculate the temperature at which \(\mathrm{rms}\) velocity of \(\mathrm{SO}_{2}\) molecules is the same as that of \(\mathrm{O}_{2}\) molecules at \(27^{\circ} \mathrm{C}\). Molecular weights of Oxygen and \(\mathrm{SO}_{2}\) are \(32 \mathrm{~g}\) and \(64 \mathrm{~g}\) respectively (A) \(327^{\circ} \mathrm{C}\) (B) \(327 \mathrm{~K}\) (C) \(127^{\circ} \mathrm{C}\) (D) \(227^{\circ} \mathrm{C}\)

At a given volume and temperature the pressure of gas (A) Varies inversely as the square of its mass (B) Varies inversely as its mass (C) is independent of its mass (D) Varies linearly as its mass

The average kinetic energy per molecule of a gas at \(-23^{\circ} \mathrm{C}\) and \(75 \mathrm{~cm}\) pressure is \(5 \times 10^{-14}\) erg for \(\mathrm{H}_{2}\). The mean kinetic energy per molecule of the \(\mathrm{O}_{2}\) at \(227^{\circ} \mathrm{C}\) and \(150 \mathrm{~cm}\) pressure will be (A) \(80 \times 10^{-14} \mathrm{erg}\) (B) \(10 \times 10^{-14}\) erg (C) \(20 \times 10^{-14}\) erg (D) \(40 \times 10^{-14}\) erg

The relation between the gas pressure \(\mathrm{P}\) and average kinetic energy per unit volume \(E\) is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\) (B) \(P=(3 / 2) E\) (C) \(\mathrm{P}=\mathrm{E}\) (D) \(\mathrm{P}=(\mathrm{E} / 2)\)
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