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  • 17. 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be <img alt="\mathrm{T_1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BT_

17. 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be \mathrm{T_1}, the work done in the process is

Option: 1

\mathrm{\frac{9}{8} R T_1}


Option: 2

\mathrm{\frac{3}{2} R T_1}


Option: 3

\mathrm{\frac{15}{8} R T_1}


Option: 4

\mathrm{\frac{9}{2} R T_1}


Answers (1)

best_answer

At STP,
22.4 L of any gas is 1 mol,

\mathrm{\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \mathrm{~mol}=n}

In adiabatic process,

\mathrm{\begin{aligned} & T V^{\gamma-1}=\text { constant } \\ & \therefore \quad T_2 V_2^{\gamma-1}=T_1 V_1^{\gamma-1} \end{aligned}}

\mathrm{\text { or } \quad T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}}

\mathrm{\gamma=\frac{C_P}{C_V}=\frac{5}{3} \text { for monoatomic He gas }}

\mathrm{\therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1}

Further in adiabatic process,
Q=0
∴ W + ?U = 0

\mathrm{\text { or } W=-\Delta U=-n C_V \Delta T}

\mathrm{\begin{aligned} &\begin{aligned} & =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\ & =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \end{aligned}\\ &=-\frac{9}{8} R T_1 \end{aligned}}

 

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Gaurav

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