One mole of an ideal gas (CP = 29.234 JK–1mol–1) is expanded reversibly and adiabatically from 1dm3 to 10 dm3. The initial temperature is 750 K, the final temperature will be:a)1000 Kb)750 Kc)300 Kd)100 KCorrect answer is option 'C'. Can you explain this answer?

Question

One mole of an ideal gas (CP = 29.234 JK&#8211;1mol&#8211;1) is expanded reversibly and adiabatically from 1dm3 to 10 dm3. The initial temperature is 750 K, the final temperature will be:a)1000 Kb)750 Kc)300 Kd)100 KCorrect answer is option 'C'. Can you explain this answer?

Answer

(T1/T2)=(v2/v1)^(y-1) [where y=Cp/Cv] Cv=Cp-R
taking log and putting the values, i got 305K
and the ans is 300K.

Answer

Reversible and Adiabatic Expansion of Ideal Gas

Given:
- CP = 29.234 JK1mol1
- Initial volume (Vi) = 1 dm3
- Final volume (Vf) = 10 dm3
- Initial temperature (Ti) = 750 K

To find: Final temperature (Tf)

Formula:
For reversible and adiabatic expansion of an ideal gas, we use the following formula:

PVγ = constant

where,
P = pressure
V = volume
γ = CP/CV (ratio of specific heat capacities)
CV = CP - R (where R is the gas constant)

Using the ideal gas equation, PV = nRT, we can rewrite the above formula as:

T(Vγ-1) = constant

where,
n = number of moles
R = gas constant

We can use this formula to find the final temperature (Tf) by using the initial conditions (Vi, Ti) and final volume (Vf).

Solution:

Step 1: Find the gas constant (R)
We can use the ideal gas equation to find the gas constant (R):

PV = nRT

R = PV/nT

Step 2: Find the constant in the formula
We can find the constant in the formula by using the initial conditions:

Ti(Viγ-1) = Tf(Vfγ-1)

Tf = Ti(Viγ-1)/(Vfγ-1)

Step 3: Find γ
We are given CP, but we need to find CV to find γ:

CV = CP - R

γ = CP/CV

Step 4: Find CV
We can use the formula for the molar heat capacity of an ideal gas:

CP - CV = R

CV = CP - R

Step 5: Substitute values and solve
Substituting the values in the formula for Tf, we get:

Tf = (750 K)(1 dm3)^(29.234 J/K/mol-1-1)/(10 dm3)^(29.234 J/K/mol-1-1)

Calculating this, we get:

Tf = 300 K

Therefore, the final temperature will be 300 K, which is option (c).

Answer

T1/T2=(V2/V1) ^y-1
Cp-Cv=R from here find the value of Cv which will come out as 20.92j/k/mol
then we know
Y=Cp/Cv
gamma value will come out as 1.4 approx
now use the above formula you will get T2 as 298.5K ~300K aprrox

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