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In the isothermal expansion of 10g of gas from volume V to 2V the work done by the gas is 575J. What is the root mean square speed ( in m/s) of the molecules of the gas at that temperature ?
Option: 1
499

Option: 2
520

Option: 3
398

Option: 4
532

Answers (1)

best_answer

$\\\text {mass of the gas} \ \ M=10 g=10 \times 10^{-3} g=\frac{1}{100} k g$ \\ $V_{1}=V: V_{2}=2 V$ \\ so therfmal process, $\Rightarrow PV=$ Constant $(c)$ \\ Work done $=W=575$

$As \ \ W=c \ln \left(\frac{V_{2}}{V_{1}}\right)=c \ln \left(\frac{2 V}{V}\right)=c \ln 2$

Comparing we get

$\begin{aligned} &\begin{array}{l} W=c \ln 2=575 \\ \Rightarrow c=\left(\frac{575}{\ln 2}\right) \ldots . .(1) \end{array}\\ &\Rightarrow PV=c=\left(\frac{575}{\ln 2}\right) \end{aligned}$

$\\ \text{As we know}, \\ \text { Root mean sq. speed }=v_{r m s}=\sqrt{\frac{3 PV}{m}}$\\ $V_{r m s}=\sqrt{\frac{3 \times 575}{(\ln 2) \times \frac{1}{100}}}=\sqrt{\frac{575 \times 300}{\ln 2}}=499 \ m/s$

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Deependra Verma

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