Please answer carefully! P1= kPa Interstellar space far from any stars is usuall
ID: 2275489 • Letter: P
Question
Please answer carefully!
P1= kPa
Interstellar space far from any stars is usually filled with atomic hydrogen (H) at a density of 1 atom/cm3 and a very low temperature of 2.73 K. Part 1 out of 3 Determine the pressure in interstellar space. At Party City, you purchase a helium-filled balloon with a diameter of 37.0 cm at 18.9 degree C and at 1.00 atm. Part 1 out of 3 How many helium atoms are inside the balloon? N = middot 1023 What is the total mass of all the oxygen molecules in a cubic meter of air at normal temperature (25 degree C) and pressure (1.01 middot 105 Pa)? Note that air is about 21% (by volume) oxygen (molecular O2), with the remainder being primarily nitrogen (molecular N2). m = g Suppose 0.0442 mole of an ideal monatomic gas has a pressure of 1.09 atm at 275 K and is then cooled at constant pressure until its volume is halved. What is the amount of work done by the gas? W = J A closed auditorium of volume 2 middot 104 m3 is filled with 6000 people at the beginning of a show, and the air in the space is at a temperature of 293 K and a pressure of l.013 middot 105 Pa. If there were no ventilation, by how much would the temperature of the air rise during the 2.00-h show if each person metabolizes at a rate of 70.0 W? Delta T = K Two containers contain the same gas at different temperatures and pressures, as indicated in the figure. The small container has a volume of 4.08 L, and the large container has a volume of 5.70 L. The two containers are then connected to each other using a thin tube, and the pressure and temperature in both containers are allowed to equalize. If the final temperature is 375. K, what is the final pressure? Assume that the connecting tube has negligible volume and mass.Explanation / Answer
1)P=dRT wher d is density in moles / m3
so P=.1/(6.023*10^23)*10^6*8.314*2.73=3.768*10^-17Pa
2)n=PV/RT=101325*4/3*pi*(0.37/2)^3/(8.314*291.9)=6.668*10^23
3)n=PV/RT=1.01*10^5*1/(298*8.314)*0.21*32=273.94g
4)0.0442*275/2*8.314=50.52 J
5)70*2*3600*6000/(1.013*10^5*2*10^4/(8.314*293)*0.02897*718)=174.8 kelvin
6)using P1V1/T1+P2V2/T2=PV/T
we get P=5.132*10^5 Pa