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Chapter 9: Problem 1249

A gas at \(27^{\circ} \mathrm{C}\) has a volume \(\mathrm{V}\) and pressure \(\mathrm{P}\). On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) \(1527^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(162^{\circ} \mathrm{C}\) (D) \(1800^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The resulting temperature of the gas is approximately \(1527^\circ C\). Therefore, the correct answer is: (A) \(1527^{\circ} \mathrm{C}\).

Step by step solution

01

Convert the initial temperature to Kelvin

We need to convert the given temperature from Celsius to Kelvin using the following formula: \(T_K = T_C + 273.15\) Given initial temperature: \(T_1 = 27^\circ C\) \(T_{1K} = 27 + 273.15 = 300.15K\)
02

Apply the combined gas law

The combined gas law formula is as follows: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We will use this formula to find the final temperature (\(T_2\)) of the gas. We know that the pressure doubles, so \(P_2 = 2P_1\), and the volume triples, so \(V_2 = 3V_1\). Plugging in these values, we get: \(\frac{P_1V_1}{300.15} = \frac{(2P_1)(3V_1)}{T_2}\)
03

Solve for the final temperature

Notice that the initial pressure and initial volume terms (\(P_1\) and \(V_1\)) will cancel out: \(\frac{1}{300.15} = \frac{6}{T_2}\) Now, rearrange the equation to solve for \(T_2\): \(T_2 = \frac{6}{\frac{1}{300.15}}\) \(T_2 = 6 \times 300.15 = 1800.9K\)
04

Convert the final temperature back to Celsius

Finally, we need to convert the final temperature from Kelvin back to Celsius: \(T_{2C} = T_{2K} - 273.15\) \(T_{2C} = 1800.9 - 273.15 = 1527.75^\circ C\) The resulting temperature of the gas is approximately \(1527^\circ C\). Therefore, the correct answer is: (A) \(1527^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

A vessel contains 1 mole of \(\mathrm{O}_{2}\) gas (molar mass 32 ) at a temperature \(\mathrm{T}\). The pressure of the gas is \(\mathrm{P}\). An identical vessel containing one mole of He gas (molar mass 4) at temperature \(2 \mathrm{~T}\) has a pressure of (A) \(2 \mathrm{P}\) (B) \(\mathrm{P}\) (C) \(8 \mathrm{P}\) (D) \((\mathrm{P} / 8)\)

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\), then percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is (A) \(33 \%\) (B) \(11 \%\) (C) \(15.5 \%\) (D) \(37 \%\)

At room temperature \(\left(27^{\circ} \mathrm{C}\right)\), the rms speed of the molecules of certain diatomic gas is found to be $1930 \mathrm{~m} / \mathrm{s}$. The gas is (A) \(\mathrm{O}_{2}\) (B) \(\mathrm{Cl}_{2}\) (C) \(\mathrm{H}_{2}\) (D) \(\mathrm{F}_{2}\)

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 molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these
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