? asked in Science & MathematicsPhysics · 1 decade ago

In a constant-volume process, 208 J of energy is transferred by heat to 0.96 mol of an ideal monatomic gas ini?

In a constant-volume process, 208 J of energy is transferred by heat to 0.96 mol of an ideal monatomic gas initially at 293 K.

(a) Find the increase in internal energy of the gas.

in J

(b) Find the work done on it.

in J

(c) Find its final temperature.

in K

Please show me the steps(formulas) as well. Thanks

3 Answers

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  • 1 decade ago
    Favorite Answer

    (a)

    The change of internal energy equals the energy transferred to the gas by work and heat.

    ∆U = W + Q

    In a constant volume process no work is done. Therefore:

    ∆U = Q = 208J

    (b)

    As point out in (a)

    W = 0

    (c)

    Internal energy of an ideal gas is given by:

    U = n∙Cv∙T

    Molar heat capacity at constant volume for an monatomic ideal gas is:

    Cv = (5/3)∙R

    Hence,

    ∆U = n∙(5/3)∙R∙∆T = n∙(5/3)∙R∙(T₂ - T₁)

    =>

    T₂ = T₁ + 3∙∆U/(5∙n∙R)

    = 293K + 3∙208J/(5 ∙ 0.96 ∙ 8.314472J/molK)

    = 309K

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  • 3 years ago

    Constant Volume Process

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    1 decade ago

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