Question

what is the change in entropy when 1 mole of helium gas is heated fom 200k to 400k at a constant pressure(given; Cp for helium =5.0 cal per degree per mole)

Answer 1

Answer:

The change in entropy for 1 mole of helium gas is equal to 1.386 cal/°C.

Explanation:

Given, the number of moles of helium gas, n = 1

The initial temperature of the gas, T₁ = 200K

The final temperature of the helium gas, T₂ = 400K

Heat capacity at constant pressure, $C_p= 5\;cal/mol \;deg$

To find the entropy change, we have formula:-

$\triangle S=nC_p\;ln \frac{T_2}{T_1}$

Substitute the value of temperatures, moles and heat capacity in above equation:-

$\triangle S=(1mol)(5\; cal \; deg^{-1}mol^{-1})\;ln \;\frac{400K}{200K}$

$\triangle S=(2\; cal\; deg^{-1})\; ln2$

$\triangle S=(2)(0.693)cal/deg$

ΔS = 1.386 cal/°C

Therefore, the change in entropy when 1 mole of helium gas is heated from 200K to 400K at a constant pressure is equal to 1.386 cal/°C.

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Answer 2

Answer:

The change in entropy of helium gas is 0.052J/K.

Explanation:

Given that,

Number of moles of helium gas is $n=1$

Initial temperature is $T_{1}=200K$ and

Final temperature is $T_{2}=400K$

Specific heat at constant pressure is

$C_{p}=5\:cal/^{0}C/mol=5(\frac{4.2J}{274K} )\\=0.076J/K/mol$

The change in entropy of the gas at constant pressure is given by the relation as shown below

$\triangle S=nC_{p}ln(\frac{T_{2}}{T_{1}})$

Substituting the given values in above relation we get

$\triangle S=1(0.076)ln(\frac{400}{200})=0.076ln2\\=0.076\times0.693J/K\\=0.052J/K$

Therefore,

The change in entropy of helium gas is 0.052J/K.

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