A gas at a volume of V\(_{0}\) in a container at pressure p\(_{0}\) is compressed to one-fifth of its volume. What will be its pressure if the magnitude of its original temperature T is constant?
\(\frac{p_{0}} {5}\)
\(\frac{4p_{0}}{5}\)
p\(_{0}\)
5P\(_{0}\)
Correct answer is D
p\(_{1}\) = P\(_{0}\), V\(_{1}\) = V\(_{0}\), V\(_{2}\) = \(\frac{1}{5}\)V\(_{0}\) P\(_{2}\) = ?
P\(_{1}\)V\(_{1}\) = P\(_{2}\)V\(_{2}\)
P\(_{2}\) = \(\frac{P_{1}V_{1}}{V_{1}}\)
= \(\frac{\not P_{o}V_{o}}{\frac{1}{5}P_{o}}\) = 5\(\not P_{o}\)