# How to convert ideal gas constant R, let say 8.31J/mol.deg to J/mol.K? ?

please help.

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?

### 2 Answers

- 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

Therefore,

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

Therefore

W = 8.31 * 293.15 * ln(20)

Follow like this.

- Login to reply the answers

- harboughLv 43 years ago
Gas Constant Conversion

Source(s): https://owly.im/a0HQC- Login to reply the answers