Heat And Thermodynamics Ques 13
- Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are $10^5 \mathrm{~N} / \mathrm{m}^2$ and $6 \mathrm{~L}$ respectively. The final volume of the gas is $2 \mathrm{~L}$ molar specific heat of the gas at constant volume is $3 R / 2$.
(1982, 8M)
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Answer:
Correct Answer: 13.( -972 J )
Solution:
Formula:
- In adiabatic process
$ \begin{aligned} p_i V_i^Y & =p_f V_f^Y \\ p_f & =\left(\frac{V_i}{V_f}\right)^\gamma p_i \end{aligned} $
$ \begin{array}{ll} \text { Further, } & C_V=\frac{3 R}{2} \\ \therefore & C_p=C_V+R=\frac{5 R}{2} \end{array} $
$ \begin{gathered} \text { or } \quad \gamma=\frac{C_p}{C_V}=\frac{5}{3} \\ \therefore \quad p_f=\left(\frac{6}{2}\right)^{5 / 3}(10)^5=6.24 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \end{gathered} $
Now, work done in adiabatic process is given by
$ \begin{aligned} W & =\frac{p_1 V_i-p_f V_f}{\gamma-1} \\ & =\frac{10^5 \times 6 \times 10^{-3}-6.24 \times 10^5 \times 2 \times 10^{-3}}{\left(\frac{5}{3}\right)-1} \\ & =-972 \mathrm{~J} \end{aligned} $
NOTE Work done is negative because volume of the gas is decreasing.
BETA
