New conversions: 1L.atm =101.325 Joules 1eV=1.6x1019 Joules R is often used in t

Question

New conversions: 1L.atm =101.325 Joules 1eV=1.6x1019 Joules R is often used in two different units systems: R 8.314J/(mol K) 0.08206 L atm/(mol K) 13 For a free expansion, you allow a volume to expand into a vacuum without doing work. simplest way to do this is to start with a gas in a volume Vo, and them release a stopper so that the volume can double (see figure). For this case no work is done on/by the system, since the space is doubled without being forced to do so by the expanding gas. Th Stopcock Given an ideal gas at Vo moles. 0.5 liters, To-300 K, n-0.5 Vacuum (a) Obtain the initial pressure Po, in atm units. (Use theI fact that the universal gas constant can be written two ways: R = 8.314 J/mol-K = 0.0821 L. atm/mol-K) Insulation (b) If the volume is allowed to double, obtain the final temperature and pressure of the system. Assume that no thermal energy enters or leaves the system 14 A typical car tire has a volume V 1.76x102 m3. At 0 Celcius the pressure inside is 2.66x105 Pascals. By driving the car, the temperature of the tire warms up to a new temperature of 30° C, and a new volume of 1.84x102 m3. Calculate the new pressure of the tire. 15 You have a mix of argon and helium atoms. Treating the system as an ideal gas, fora temperature of T 100 Kelvin, (a) calculate the root mean square speed of the argon atoms. (b) calculate the root mean square speed of the helium atoms. (The molar mass of argon is 40 g/mol and the molar mass of helium is 4.0 g/mol

Explanation / Answer

14) Given,

V1 = 1.76 x 10^-2 m^3 ; P = 2.66 x 10^5 Pa ; T1 = 0 deg = 273 K ; T2 = 30 deg C = 303 K

assuming that the gas inside tire is ideal gas. We know for an ideal gas:

P1V1/T1 = P2V2/T2

P2 = P1 V1 T2/V2 T1

P2 = 2.66 x 10^5 x 1.76 x 10^-2 x 303/(1.84 x 10^-2) x 273= 2.82 x 10^5 Pa

Hence, P2 = 2.82 x 10^5 Pa

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