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Chapter 13: Problem 43

Verify, using the ideal gas law, the assertion in Problem 38 that 1.00 mol of a gas at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm occupies a volume of $0.0224 \mathrm{m}^{3}$

Short Answer

Expert verified
Question: Verify the following assertion using the ideal gas law: 1.00 mole of an ideal gas at 0.0 °C and a pressure of 1.00 atmosphere occupies a volume of 0.0224 m³. Answer: The assertion is verified using the ideal gas law. The calculated volume of the gas is 0.0224 m³, which matches the given assertion.

Step by step solution

01

Convert the temperature to Kelvin

Convert the given temperature in Celsius to Kelvin: T(K) = T(°C) + 273.15, where T(K) is the temperature in Kelvin and T(°C) is the temperature in Celsius. T(K) = 0.0 + 273.15 = 273.15 K
02

Convert the pressure to Pascals

Convert the given pressure in atmospheres to Pascals: P(Pa) = P(atm) * 101325, where P(Pa) is the pressure in Pascals and P(atm) is the pressure in atmospheres. P(Pa) = 1.00 * 101325 = 101325 Pa
03

Substitute the values into the ideal gas law equation

Substitute the given values and the converted values of temperature and pressure in the ideal gas law equation: PV = nRT (101325 Pa)(V) = (1.00 mol)(8.314 J/(mol·K))(273.15 K)
04

Solve for the volume of the gas

Solve the equation for volume (V) by dividing both sides by 101325 Pa: V = (1.00 mol)(8.314 J/(mol·K))(273.15 K) / 101325 Pa = 0.0224 m³ The calculated volume of the gas is 0.0224 m³, which matches the assertion given in the problem. Therefore, the assertion is verified using the ideal gas law.

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Most popular questions from this chapter

The volume of a solid cube with side \(s_{0}\) at temperature \(T_{0}\) is \(V_{0}=s_{0}^{3} .\) Show that if \(\Delta s < s_{0},\) the change in volume \(\Delta V\) due to a change in temperature \(\Delta T\) is given by $$ \frac{\Delta V}{V_{0}}=3 \alpha \Delta T $$ and therefore that \(\beta=3 \alpha .\) (Although we derive this relation for a cube, it applies to a solid of any shape.)
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