Calculations: Show all your work in solving the following calculations, label al
ID: 578126 • Letter: C
Question
Calculations: Show all your work in solving the following calculations, label all answers with appropriate unit 33. A container of oxygen has a volume of 30.0 mL and a pressure of 4.00 atm. The pressure of the oxy the oxygen gas, report your answer in liters. gen gas is reduced to 2.88 atm while the temperature is kept constant. Calculate the new volume of at 8 )0.04 120-2-88Vz-1.666 : 2.88atm 1.7 34. How much energy is required to raise the temperature of a 1.88-g solid sample of iron from 15.0°C to 35.0 °C? Assume specific heat of iron, 0.45 Jig cc. 35. A sample of argon gas occupies 11.5 L at 25 °C and 1.57 atm. What volume will it occupy at STP?Explanation / Answer
Gas law equation can be written as
PV= nRT
R = universal gas constant and n= no of moles
P= Pressure , V= Volume and T= temperature in K
For a given mass of gas at constant temperature,
The equation becomes
PV= constant or at two different conditions of Pressure (P1 and P2)
And corresponding volumes V1 and V2
P1V1= P2V2
Given P1= 4 atm, V1= 30ml, P2= 2.88 atm
V2= P1V1/P2= 4*30/2.88 =41.7 ml
2.
. Energy required = mass of iron* specific heat of iron * temperature change
=1.88 gm*0.45 J/gm.deg.c*(35-15)= 16.92 joules
3.
Gas law equation can be written as
PV= nRT, R = universal gas constant and n= no of moles, P= Pressure , V= Volume and T= temperature in K
At two different conditions of temperature, pressure and Volume
The equation becomes
P1V1/T1= P2V2/T2, P2, V2 and T2 refer to standard conditions of pressure, temperature and volume.
Where P1= pressure , V1= volume , T1= temperature in K
Given P1= 1.57 atm, V1= 11.5L, T1= 25 deg.c= 25+273= 298K
P2 =1 atm , T2= 273.15K (STP conditions )
V2= P1V1T2/P2T1= 1.57* 11.5*273.15/(1*298)=16.55 L
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