Question
Question: A mixture of gases at NTP for which \(\Upsilon = 1.5\) is suddenly compressed to \(\dfrac{1}{9}th\) ...
A mixture of gases at NTP for which Υ=1.5 is suddenly compressed to 91th of its original volume. The final temperature of mixture is:
A. 300∘C
B. 546∘C
C. 420∘C
D. 872∘C
Solution
Hint: Remember to consider the initial temperature of the mixture as 273K also use the given statement to find the final volume of the mixture and use the information to find the final temperature of the mixture.
Complete answer:
We know Initial temperature (T1) = 0°C = 273 K
Let Initial volume (V1) = V
Final volume (V2) = V/9
Υ=1.5
According to adiabatic process, PVΥ= Constant
Since P = T/V
Therefore VTVΥ−1=c
TVΥ−1=c
For initial condition T1V1Υ−1=c (equation 1)
For final condition T2V2Υ−1=c (equation 2)
Equating equation 1 and 2
T1V1Υ−1=T2V2Υ−1
According to question V2=V191
Therefore T2=T1(V2V1)Υ−1
⇒T1(9)0.5
⇒3×273
⇒819K=546∘C
Therefore the final temperature of the mixture is 546∘C.
Note: The concept of Adiabatic process is defined as the thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression. Adiabatic processes can be reversible or irreversible. But for adiabatic processes to occur there are some conditions in which an adiabatic process can take place is that the system must be perfectly insulated from the surrounding and also the process must be carried out quickly so that there is sufficient amount of time for heat transfer to take place.
The Adiabatic process Equation is PVΥ= Constant here P is the pressure of the system, V is the volume of the system and Υis the adiabatic index which is defend as the ratio of heat capacity at constant pressure CP to heat capacity at constant volume CV.