such as chlorine (Cla) (6) 573 4) 825 room temperature 5. Which of the following
ID: 1043459 • Letter: S
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such as chlorine (Cla) (6) 573 4) 825 room temperature 5. Which of the following relationships between the two variables listed for an ideal gas does not give a straight lie graph with a positive slope? (a) P and T (b) n and Vv (c) P and V (d) T and V (e) none of these (e) SO the parient breathed 0.350L efO Three containers of equal volume are filled with three different gases. The first container is illed with A, the second with O and the third with COs. The pressure and temperature of the three gases are the same. Which of tde following is true of the gases in the three containers? ()0.006131 (b) 0.244 1 (c) 0250 L (4) 0.2561 (e) none of the above (a) The number of moles of gas in the containers is the same (b) The masses of gas in the three containers are the same. (c) The density of gas in the three containers is the same (d) (a), (b), and (c) are all true. (c) None of these are true. 23. Which of the following can explain the increse in r that results from decreasing the volume of s fxed mont of gas at a fixed temperature? (a) The gas molecules move faster and therefore hit the The density of neon gas will be greatest at walls of the container with more force. (b) The gas molecules have a shorter path between the walls of the container and therefore collide with the (a) 0°C and I atm (b) 273 °C and 1.5 atm walls more frequently. (c) The gas molecules increase in their apparent mass C and 2 atm therefore exerting more force when they strike the a) 273°C and 1 atm walls of the container (d) The gas molecules collide less with each other therefore (e) The density of neon will be the same for all of the above cases having more force to exert on the walls of the (e) None of these. Calculate the volume in liters occupied by 2.0 g of Ne at Which would always lead to an increase in the average kinetic energy of a gas? 50.0°C and 1.5 atm of pressure. 24. (b) 8.8 L (c) 35 L (d) 47 L e) none of these (a) Increasing the volume by decreasing the pressure. (b) Increasing the pressure by decreasing the volume (c) Increasing the pressure by increasing the number of molecules of gas. (d) Increasing the volume by increasing the temperature o the gas. (e) All of the above are equally effective ways of increasi 2.91 gram sample of a gaseous compound that contains d 1.09 atm. What is the formula of this compound? B2H6 nly boron and hydrogen has a volume of 1.22 L at 25 C the average kinetic energy of a gas. 25. Which of the following gases woulddiffis fasts t roos 10 BsHo (a) NH (b) CO (c) HS (d) F2 (e) CO 10 12 t is the final temperature in kelvins if a sample of onia gas, initially at a pressure of 3.00 atmospheres, a erature of 500.0 K, and a volume of 275 L is changed olume of 200.0 L and a pressure of 2.50 atm?Explanation / Answer
Gas law equation is PV= nRT
at constant mass and volume of gas , P= KT, K is constant the plot of P Vs T are straight line
At constant pressure and temperature, V= K*n, K is cinstant and hence the plot of V vs n is straight line.
at constant pressure for a given mass of gas , V= KT, K is constant and hence V Vs T is straight line.
At constant temperature for a given mass of gas, PV =nRT becomoes PV= K, a constant. So a plot of P vs V cannnot be straight line. ( C is correct)
2.from gas law, PV= nRT, n= no of moles= PV/RT, at constant pressure, volume and temperature, n is same for all the gases. But n = mass/molar mass, mass= n* molar mass, since molar mass is different for the gases, mass is not same for the gases. since density= mass/volume and mass is different, density is also different. ( So A is correct)
3. PV= nRT, n= mass/molar mass
PV= (mass/molar mass)*RT
P* molar mass= (mass/volume)*RT
P* molar mass= density*RT
Density= P* molar mass/RT
higher the pressure and lower the temperature, higher the density. for option-C, Pressure is more and temperature is less. So correct answer.
4. Molar mass of Neon =20 g/mole, moles of Neon, n = 2/20=0.1, T= 50 deg.c= 50+273= 323K, P= 1.5 atm, R= 0.0821 L.atm/mole.K, from gas law, V = nRT/P=0.1*0.0821*323/1.5=1.76 L ( none of the options given)
5. given P= 1.09 atm, V= 1.22, T= 25 deg,c= 25+273= 298K, R= 0.0821 L.atm/mole.K
n= no of moles= PV/RT= 1.09*1.22/(0.0821*298) =0.054
moles= mass/molar mass
molar mass= mass/moles= 2.91/0.054=54 g/mole
atomic weight of B= 11 and H= 1 ,so empricai formula is
This matches wirth B4H10. ( B is correct)
5. given P1=3 atm, V1=275L and T= 500K, P2=2.5 atm, V2=200L
from gas law, P2V2/T2= P1V1/T1
T2= P2V2T1/P1V1= 2.5*200* 500/(3*275)=303K
6. moles= mass/ molar mass, mole fraction = moles of gas/total moles, partial pressure= mole fraction* total pressure,. since total moles of gas is same, which ever gas is having more no of moles, will have more mole fraction and hence more parital pressure
molar masses ( g/mole): H2= 2, CH4=16, SO2=64 and Ar= 40
moles =mass/molar mass, moles : H2= 10/2=5, CH4=15/16=0.9375, SO2= 20/64=0.3125 and Ar= 12/40=0.3. Hence hydrogen will have more partial pressure.
7.
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