# 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

Relevance

(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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