Answer:
a) 
b) 
d) 
d) 
Explanation:
From the question we are told that:
Moles of N2 
Atmospheric pressure 
Temperature 

Initial heat 
a)
Generally the equation for change in temperature is mathematically given by

Where




b)
Generally the equation for ideal gas is mathematically given by

For v double


Therefore



Total Work-done 



c)
Generally the equation for amount of heat added is mathematically given by



d)
Generally the equation for change in internal energy of the gas is mathematically given by



Answer:
2.47L
Explanation:
Using the combined gas law equation as follows:
P1V1/T1= P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 705mmHg
P2 = 760mmHg (STP)
V1 = 3.00L
V2 = ?
T1 = 35°C = 35 + 273 = 308K
T2 = 273K (STP)
Using P1V1/T1= P2V2/T2
705 × 3/308 = 760 × V2/273
2115/308 = 760V2/273
Cross multiply
308 × 760V2 = 2115 × 273
234,080V2 = 577,395
V2 = 577,395 ÷ 234,080
V2 = 2.47L
The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
Answer:
Because Iodine is a non-polar but water is polar therefore it can not have a hydrogen bond or any permanent dipole-dipole interactions
Noble gases are known for having a full outer shell of electrons which helium has as it has two electrons its first electron shell is completely filled