Question

When an evacuated tube of volume 400 cm3 is filled with gas at 300K and 101 kPa, the mass of the tube increases by 0.65 g. Assume the gas behaves as an ideal gas. What could be the identity of the gas? Show your working

Answer 1

Explanation:

P volume = 0.112 L

volume = 0.112 Lpressure = 3.75 atm

volume = 0.112 Lpressure = 3.75 atmtemp = 273K

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRT

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn=

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×273

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 moles

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measured

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20 Hence, the correct option is C

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20 Hence, the correct option is CAnswered By

Answer 2

Given:

Temperature, T = 300 K

Volume, V= 400cm³

Pressure, P= 101kPa = 101 × 10³Pa

Weight, W= 0.65g

To Find:

Identity of the gas

Solution:

The gas behaves as an ideal gas,

Hence,

PV = nRT = WRT/M

M = WRT/PV

= 0.65×8.314×300/101×10³×4×10^(-4)

=4.01×10

=40

Hence, the answer is 40.

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