Answer
a) $100\;\rm nm$
b) $400\;\rm nm$
Work Step by Step
a)
We know that the mean free path of a gas is inversely proportional to its pressure, so
$$\lambda\propto \dfrac{1}{P}$$
Hence,
$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{P_2}{P_1}$$
If the pressure of the gas is doubled while all other state variables are held constant, the mean free path of molecules in the gas will decrease. v
$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{2P_1}{P_1}$$
Thus,
$$\lambda_2=\frac{1}{2}\lambda_1=\frac{1}{2}(200)=\color{red}{\bf 100}\;\rm nm$$
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b)
We know that the mean free path of a gas is directly proportional to its temperature, so
$$\lambda\propto T$$
$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{T_1}{T_2}$$
$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{T_1}{2T_1}$$
Hence,
$$\lambda_2=2\lambda_1=2(200)=\color{red}{\bf 400}\;\rm nm$$
