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Chapter 5: Problem 67

An ideal gas is contained in a cylinder with a volume of \(5.0 \times\) $10^{2} \mathrm{mL}\( at a temperature of \)30 .^{\circ} \mathrm{C}$ and a pressure of 710 . torr. The gas is then compressed to a volume of 25 \(\mathrm{mL}\) , and the temperature is raised to \(820 .^{\circ} \mathrm{C}\) . What is the new pressure of the gas?

Short Answer

Expert verified
The new pressure of the gas is approximately 1706 torr.

Step by step solution

01

Write down the initial and final conditions of the gas.

The initial and final conditions of the gas are as follows: Initial Condition Volume, V1 = \(5.0 \times 10^2 \mathrm{mL}\) Temperature, T1 = \(30 .^{\circ} \mathrm{C}\) Pressure, P1 = 710 torr Final Condition Volume, V2 = 25 \(\mathrm{mL}\) Temperature, T2 = \(820 .^{\circ} \mathrm{C}\) Pressure, P2 = ? With these conditions, we can now apply the ideal gas law formula to solve for the new pressure, P2.
02

Convert temperature to kelvin and pressure to atmosphere.

Before applying the ideal gas law, standardize the temperature unit, and convert the pressure unit (from torr to atmospheres): T1 = \(30.^{\circ} \mathrm{C}\) + 273.15 = 303.15 K T2 = \(820 .^{\circ} \mathrm{C}\) + 273.15 = 1093.15 K 1 atm = 760 torr P1 = 710 torr * \( \frac{1 \; \mathrm{atm}}{760 \; \mathrm{torr}} \) = \( \frac{710}{760} \) atm
03

Use the ideal gas law to eliminate moles and arrive at combined gas law.

We will use the ideal gas law PV = nRT, and the fact that the number of moles stays the same to arrive at the combined gas equation for the two states of the gas in terms of pressure, volume, and temperature: Since P1V1/n1T1 = P2V2/n2T2 and n1=n2, then: \( \frac{P1 \cdot V1}{T1} \) = \( \frac{P2 \cdot V2}{T2} \)
04

Solve for the new pressure (P2).

With the combined gas equation in hand, plug in the values for P1, V1, V2, T1, and T2: \( \frac{(\frac{710}{760}\;\mathrm{atm})(5.0 \times 10^2 \; \mathrm{mL})}{303.15 \; \mathrm{K}} \) = \( \frac{P2(25\; \mathrm{mL})}{1093.15\; \mathrm{K}} \) Rearrange the equation and solve for P2: P2 = \( \frac{(\frac{710}{760}\; \mathrm{atm})(5.0 \times 10^2 \; \mathrm{mL})(1093.15\; \mathrm{K})}{303.15 \; \mathrm{K} (25\; \mathrm{mL})} \) P2 ≈ \((2.243 \; \mathrm{atm})\)
05

Convert the new pressure to torr.

Now, convert the new pressure back to torr to obtain the final answer: P2 = 2.243 atm * \( \frac{760 \; \mathrm{torr}}{1\; \mathrm{atm}} \) ≈ 1705.64 torr The new pressure of the gas is approximately 1706 torr.

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