The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be

Question

KollegeApply · Accepted Answer

100 kPa Let the number of moles in the gas be n. Applying equation of state, we get PV = nRT \Rightarrow P = nRT/V \Rightarrow 2 $\times$ {10}^5 = nRT/V . . . \left( 1 \right) \text { When half of the gas is removed, number of moles left behind } = n/2 \text { Let the pressure be P' }. P' = n/2RT/V \text { Now }, P' = $\frac{1}{2}$ $\times$ 2 $\times$ {10}^5 = {10}^5 \left[ \text { From eq } . \left( 1 \right) \right] =100 kPa

AChoose Your Answer