A student was asked to convert 720 torr to atm. He wrote the following formula.
ID: 1009378 • Letter: A
Question
A student was asked to convert 720 torr to atm. He wrote the following formula. Is there anything wrong^3 atm = 720 torr x 760 terr/1 atm = some value Is this the right way of writing? If not, write the right conversion and write the value in atm. A 4.00 L balloon it purchased at a gift shop at 23 degree C. If the balloon shrinks to 3.45 I, when taken outside: To find the temperature outside, what gas-law do you apply? What is the right formula you apply? What is temperature outside in kelvin? If the pressure is different inside and outside the gift shop, do you still apply the same formula? If not, explain why? Can you write the right formula if pressure is 1 atm inside the gift shop and 1.3 atm outside gift shop? What will be the new value of outside temperature then? Consider 2.0 I, helium m a balloon at 75 degree F and at atmospheric pressure. What formula do you apply to solve the number of moles (n)? A student solved this problem by using the value of temperature as 75 degree F as such for T. Did she use right value? If not, explain why? What is R in the formula you wrote in question-1? What are the units of R? If you apply this formula, what will be the value of number of moles of helium gas?Explanation / Answer
1)760 torr= 1 atm
720 torr= 720(torr)* 1 atm/ 760 torr= 0.947 atm
2) Charless law is used which states that for a given mass of gas at constant pressure, VT= Constant
V1/T1= V2/T2
T2= V2*T1/V1
V1= 4 L T1= 23+273.15= 296.15 K V2=3.45 L T2= 3.45* 296.15/4=255.43 K
when pressure changes, the gas law equation will have to be used.
P1V1/T1= P2V2/T2
T2= P2V2T1/P1V1= 1.3*3.45*296.15/4*1= 332.1
3. V= 2 L t= 75 deg.F= (75-32)/1.8 deg.c=24deg.c= 24+273.15= 297.15K P= 1 atm
PV= nRT will have to be used, R=0.0821L.atm/mole.K
n= PV/RT= 1*2/ (0.0821*297.15)= 0.082 moles of helium
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