Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
![R\propto \sqrt{\frac{1}{M}}](https://tex.z-dn.net/?f=R%5Cpropto%20%5Csqrt%7B%5Cfrac%7B1%7D%7BM%7D%7D)
or,
..........(1)
where,
= rate of effusion of nitrogen gas = ![79mL/s](https://tex.z-dn.net/?f=79mL%2Fs)
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
![(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}](https://tex.z-dn.net/?f=%28%5Cfrac%7B79mL%2Fs%7D%7BR_2%7D%29%3D%5Csqrt%7B%5Cfrac%7B64g%2Fmole%7D%7B28g%2Fmole%7D%7D)
![R_2=52mL/s](https://tex.z-dn.net/?f=R_2%3D52mL%2Fs)
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
Thermodynamic quantity equivalent to the total heat content of a system It is equal to the internal energy of the system plus the product of pressure and volume
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ![\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )](https://tex.z-dn.net/?f=%5Cfrac%7BH_%7Bvap%7D%7D%7Br%7D%20%28%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%20%20%29)
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:
![T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}](https://tex.z-dn.net/?f=T_%7B2%7D%3D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%20-%5Cfrac%7BRln%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%7D%7BDelta%20H_%7Bvap%7D%7D)
(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C