Question

The pressure of a gas is $PN/m^2$. The kinetic energy of one gram-mole gas at NTP (in joule) will be

• A. $1.68\times {10}^{-2}P$
• B. $1.5P$
• C. $2.24\times {10}^{-2}P$
• D. $3.36P$

Answer 1

Answer: (A)

$1.68\times {10}^{-2}P$

At NTP, pressure = $1.013\times {10}^{5}N/m^2$

In the question, it is given that $P=PN/m^2$

$\Rightarrow P=1.013\times {10}^5$

Now, we know that at NTP, $K.E=\frac{3}{2}nRT$

$\Rightarrow K.E=\frac{3}{2}\times 8.314\times 273=1702.29J$

We can write it as $K.E=\frac{1702.29}{1.013\times {10}^5}=1.68\times {10}^{-2}P$