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jai asked in Science & MathematicsPhysics · 1 decade ago

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

please help.

Update:

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

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  • 1 decade ago
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    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.

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  • 3 years ago

    Gas Constant Conversion

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