How to convert ideal gas constant R, let say 8.31J/mol.deg to J/mol.K? ?
If this are same, can I get the same answer? If let say in calculation for Work in compressing piston for 1 mol of ideal gas which is pressure are change from 20 atm to 1 atm in constant temperature, say 20°C. I used W = intergrate_PdV but at the end I could't get the correct answer. What seems to be the problem?
- AvinashLv 71 decade agoFavorite Answer
Both are same.
This means it is 8.31 J/mol.K
You may be wondering that K and deg are different. Then how can J/mol.deg be same as J/mol.K?
This is because deg or K in R is change in temperature.
Take an example. Suppose temperature of an object is 5 deg C.
You increase the temperature by 15 deg C. It becomes 20 deg C.
Initial temperature in K is 5 + 273.15 = 278.15 K
Final temperature in K is 20 + 273.15 = 293.15 K
Change in temperature = 293.15 - 278.15 = 15 K
As you can see that in deg C, the change is 15 and in K also, the change is 15.
The temperature of an object is different in K and deg C. But change is temp. is the same.
PV = nRT
n = 1
PV = RT
Or V = RT/P
Differentiating both sides,
dV = RT(-1/P^2)dP
Let work done = W
W = integral P dV
Or W = integral P RT(-1/P^2)dP
= integral RT(-1/P) dP
R and T are constants.
Therefore W = -RT * integral (1/P) dP
= -RT * ln(P) dP
Take lower limit as P1 and upper as P2.
W = -RT (ln P2 - ln P1)
Or W = RT(ln P1 - ln P2)
Or W = RT * ln(P1/P2)
Or W = RT * ln(20/1)
Or W = RT * ln(20)
R = 8.31 J/mol.K,
T = 20 deg C = 20 + 273.15 K = 293.15 K
W = 8.31 * 293.15 * ln(20)
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