In van der Waals equation at constant temperature 300 ~K , if a=1.4 ~atm ~L ~mol^ -2 , V=100 ~mL, n=1 ~mole , then what is the pressure of the gas?

Correct answer: D. 106 atm At moderate pressure van der Waals equation is given as : aligned & (P+(a n^2)/(V^2))(V)=n R T & (P+(1.4)/((0.1)^2))(0.1)=1 × 0.082 × 300 aligned aligned & (P+140) × 0.1=24.6 & 0.1 P+14=24.6 & 0.1 P=10.6 P=106 ~atm aligned

AIIMS 201827 May

Exam: AIIMS
Year: 2018
Date: 27 May
Subject: Chemistry
Chapter: States of Matter

AIIMS 2018 · 27 May · Chemistry · States of Matter

In van der Waals equation at constant temperature 300 ~K , if a=1.4...

In van der Waals equation at constant temperature

300 \mathrm{~K}

, if

a=1.4 \mathrm{~atm} \mathrm{~L} \mathrm{~mol}^{-2}

V=100 \mathrm{~mL}, n=1 \mathrm{~mole}

, then what is the pressure of the gas?

Options

A42 atm
B210 atm
C500 atm
D106 atm

Correct answer

D. 106 atm

Step-by-step solution

At moderate pressure van der Waals equation is given as :

\begin{aligned} & \left(P+\frac{a n^2}{V^2}\right)(V)=n R T \\ & \left(P+\frac{1.4}{(0.1)^2}\right)(0.1)=1 \times 0.082 \times 300\end{aligned}

\begin{aligned} & (P+140) \times 0.1=24.6 \\ & 0.1 P+14=24.6 \\ & 0.1 P=10.6 \Rightarrow P=106 \mathrm{~atm}\end{aligned}

Practice States of Matter on Quantrex Academy →

More from States of Matter