Question

Simple gas laws

What is the effect of the following changes on the volume of 1 mol of an ideal gas? (Note: You should NOT need to do any explicit numerical calculations. Use your qualitative understanding of how volume depends on different experimental variables.)

The pressure is changed from 2 atm to 760 torr (at constant temperature).

（1， 1/2， 1/4，2）

Half the gas escapes through a stopcock (at constant temperature and pressure).	Answer 2
Both the pressure and the temperature are reduced to 1/4 of their initial values.	Answer 3
The temperature is changed from 546 K to 0 °C (at constant pressure).	Answer 4
The absolute temperature is doubled and the pressure is changed from 760 mm Hg to 0.5 atm.

Answer 1

PV = nRT [ Ideal Gas equation P = Pressure of gas V = volume n = No of moles of gou R = universal Gas Constant -I = Temperature. n = 1 mole La P, = Initial Pressure = 2atm V = Initial volume = final volume P2 = final Pressure = 760tors = latma latm = 760tor PV = nRT 2 qtmx V = I molexRT - P2 V -ORT latmx V₂ = Imole XRT 2) Dividing equation 1st by equation and we get V2 ₂ = av, final volume will become twice of initial (b) Half of gas escapes no = Initial No of mp = final No of through stopcock moles of goo. moles of gae PV = nRT 1)

- PV = nRI Dividing equation ist by equation and we get Pvi PVE iRI [na vi Uf = 2 vi I final volume will become I of initial Volume 1 C Initial temperature - Ti = 546K final temperature = If = 0°c = 273K Vi = Initial volume Vf = final volume PV = nRT? = PV = nRT = nRx 546K nRX 27382 and we get Divide V equation est by equation = 546 = 2 23 u Vf = Ivi a final volume will becom half of initial.

= 760mmHg zlata - PE = 0.5 atom o Tf = final temperature = a Ti Pivo = nRTi Tamm X V1 = ORT; Pf Vf =nRTF 0.5 atm x VF =nRXaTi - 2) Dividing equation ist by equation and we get VE = 4 vi final volume will become 4 times of initial