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Two moles of an ideal gas cv-5/2 wascompressed adiabatically against constant pressure of 2atm. Which was initially at 350k and 1pressure.the work involve in the process is equal to how much

Two moles of an ideal gas cv-5/2 wascompressed adiabatically against constant pressure of 2atm. Which was initially at 350k and 1pressure.the work involve in the process is equal to how much

Grade:12

3 Answers

Sakshi
askIITians Faculty 652 Points
8 years ago
hello..just use the basics of thermodynamics for adiabatic processes..
keeping q=0
pfa
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Raunaq Mehta
39 Points
7 years ago
∆q=0 ,∆U=w. W=-pext(V2-V1) and U=ncvmdTEquating W and U u get final temperature as 450K. From there u can calculate work done and u get the answer as 500R
disha sharma
29 Points
2 years ago

For reversible adiabatic process,

 
 
W=−Pext​[P2​nRT2​​−P1​nRT1​​]              
 P2​=Pext​=2atm
 
P1​=1atm                                              T1​=300K
 
W=−(2atm)[2atm2(R).T2​​−1atm2R(350)​]
 
and W=2CV​(T2​−350)=2×25​R(T2​−350)
 
5R(T2​−350)=(750R−2RT2​)
 
5T2​−1750=1400−2T2​
 
7T2​=3150                          T2​=450K
 
W=2×CV​(450−350)
 
=2×25​R×(100)=500R
 

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