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Chapter 5: Problem 41

A gas at \(772 \mathrm{mmHg}\) and \(35.0^{\circ} \mathrm{C}\) occupies a volume of \(6.85 \mathrm{~L}\). Calculate its volume at STP.

Short Answer

Expert verified
The volume of the gas at Standard Temperature and Pressure (STP) is approximately \(6.08 \mathrm{L}\).

Step by step solution

01

Convert all values to standard units

Firstly, convert the given temperature to Kelvin by adding 273 to the Celsius value. So, \(35.0^{\circ}C\) translates to \(35+273 = 308 \mathrm{K}\). Next, convert the pressure from \(mmHg\) to \(atm\) by dividing by 760. Thus, \(772 \mathrm{mmHg}\) is equal to \(772/760 = 1.016 \mathrm{atm}\).
02

Input values into the formula

The standard conditions for gases is defined as 0 degrees Celsius or 273 K for temperature and 1 atm for pressure.Substitute \(P_1\), \(V_1\), and \(T_1\) with the initial conditions, and \(P_2\) and \(T_2\) with the STP conditions. Hence, the equation becomes \((1.016 \mathrm{atm}\) * \(6.85 \mathrm{L}\)) / \(308 \mathrm{K}\) = (1 atm * \(V_2\)) / (273 K).
03

Solve for \(V_2\)

To find the unknown \(V_2\), cross-multiply and solve for \(V_2\). You get \(V_2\) = \((1.016 \mathrm{atm} * 6.85 \mathrm{L} * 273 \mathrm{K}) \) / \((1 \mathrm{atm} * 308 \mathrm{K})\). After calculating you get \(V_2 \approx 6.08 \mathrm{L}\)

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Most popular questions from this chapter

In \(2.00 \mathrm{~min}, 29.7 \mathrm{~mL}\) of He effuse through a small hole. Under the same conditions of pressure and temperature, \(10.0 \mathrm{~mL}\) of a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) effuse through the hole in the same amount of time. Calculate the percent composition by volume of the mixture.

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