Discussion Forum

An ideal gas is placed in a tank at $27^{0} C$. The pressure is initially 600 kPa. One-fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{0}$ C?

Answer:

Explanation:

 As, we know that , the ideal gas equation is given as, 

          pV=nRT

It can written as,

 $\frac{p_{1}V_{1}}{n_{1}T_{1}}=\frac{p_{2}V_{2}}{n_{2}T_{2}}$  ........(i)    ($\because$  R= constant)

 Given, $T_{1}=273+27^{0}=300K$ and $p_{1}=600 KPa$

If one fourth of gas is released from tank,

then , $n_{2}= \frac{3}{4}n_{1}$ and   $V_{2}=V_{1}$    ($\therefore$  No change)

Given, $T_{2}$=327+273=600K

So, from Eq.(i)  , we get

$\frac{600V_{1}}{n_{1}300}=\frac{p_{2}V_{2}}{\frac{3}{4}n_{1}600}\Rightarrow p_{2}=900KPa$

 So, the correct option is (a)