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Chapter 5: Problem 94

At \(27^{\circ} \mathrm{C}, 10.0\) moles of a gas in a 1.50 -L container exert a pressure of 130 atm. Is this an ideal gas?

Short Answer

Expert verified
The comparison of calculated R value with the standard Ideal Gas constant R will tell us if the gas behaves ideally or not. If the values match, the gas is ideal; if not, the gas is non-ideal.

Step by step solution

01

Convert Celsius to Kelvin

The question gives the temperature in Celsius, but for the Ideal Gas Law calculation, we have to convert the temperature to Kelvin. To do so, you add 273.15 to the Celsius temperature. In this case, converting 27 degrees Celsius to Kelvin is done by: 27 + 273.15 = 300.15 K.
02

Calculate R using given data

Now we have to rearrange the Ideal Gas Law: \(PV=nRT\) to calculate the value of R. The rearrangement gives us: \(R = PV/nT\). Substituting P=130 atm, V=1.50 L, n=10.0 moles, and T=300.15 K, into the equation will give us: \(R = (130 * 1.50) / (10 * 300.15)\)
03

Comparison with standard R

After we compute the value of R, we compare it with the known Ideal Gas constant R = 0.0821 L.atm/K.mol. If the values match (or the difference is negligible), the gas behaves ideally. If the calculation value of R is significantly different from standard R, the gas doesn't behave ideally.

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Most popular questions from this chapter

A mixture of helium and neon gases is collected over water at \(28.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg} .\) If the partial pressure of helium is \(368 \mathrm{mmHg}\), what is the partial pressure of neon? (Vapor pressure of water at \(28^{\circ} \mathrm{C}=\) \(28.3 \mathrm{mmHg} .\) )

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