Please answer all the following questions (1-5) with work shown as neatly as pos
ID: 1604934 • Letter: P
Question
Please answer all the following questions (1-5) with work shown as neatly as possible. Thank you very much!
A building made with a steel structure is 650 m high on a winter day when the temperature is 0 degree F. How much taller (in cm) is the building when it is 100 degree F? (The linear expansion coefficient of steel is 11 times 10^-6 (degree C)^-1.) A beaker is filled to the 500 ml mark with alcohol. What increase in volume (in ml) does the beaker contain when the temperature changes from 5 degree C to 30 degree C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) beta_alcohol = 1.12 times 10^-4/degree C One mole of an ideal gas is held at a constant pressure of 1 atm. Find the change in volume (in liters) if the temperature changes by 50 degree C. A container with a one-liter capacity at 21 degree C is filled with helium to a pressure of 2.0 atm. (1 atm = 1.0 times 10^5 N/m^2.) How many moles of helium does it hold? A container having a volume of 1.0 m^3 holds 5.0 moles of helium gas at 50 degree C. If the helium behaves like an ideal gas, what is the total energy of the system?Explanation / Answer
1.
Given
height of the building made up of steel , in winter is l1 = 650 m at 0 0F
the height of the building when its temperature changed from 0 - 100 0F is l2=?
linear coefficient of expansion alpha = 11*10^-6 / 0C
we know that from definition of alpha = (l2 - l1)/(l1*(T2-T1))
l2 = l1(1+ alpha(T2-T1))
l2 = 650(1+11*10^-6(100-0)) m
l2 = 650.715 m
2. volume of beaker with alcohol is v1 = 500 ml
Temperature changed from 5 0C to 30 0C
we know that the real expansion of the liquid is gamma_r = (v2-v1)/(v1(T2-T1))
v2= v1(1+ gamma_r(T2-T1))
= 0.5(1+1.12*10^-4(30-5)) lt
= 0.5014 lt = 501.4 ml
3.
ideal gas
Pressure P = 1 atm
no of moles n = 1
temperature change dT = 50 0C
change in volume of the gas is dV = ?
from ideal gas equation PV = nRT
V = nRT/p
dV = nR*dT /P
= 1*0.082057 *50/1 lt
= 4.10285 lt
change in volume is dV = 4.10285 lt
4. volume of helium in the container is v = 1 lt
pressure P = 2 atm = 2*1.0*10^5 Pa = 2*10^5 Pa
Temperature is T = 27 0C = 300.15 k
ideal gas equation Pv = nRT
n = PV/RT
= 2*10^5*1/(8.314472*300.15)
= 80.1414
5.
no of moles n = 5 , volume is v = 1 m^3
Temperature is T = 50 0C = 50+273.15 = 323.15 k
we know that the total kinetic energy of the ideal gas is k.e_total = 3/2*n*R*T
k.e = (3/2)(5*8.314472*323.15) J
k.e = 20151 J