Determine the pressure $p_1$ on the left side of a box as a function of time after a small hole of area $A$ is created in the partition dividing a volume $2V$ into two halves. Initially, the left side contains a perfect gas of density is $n_0$ at pressure $p_0$ , while the right side is a vacuum. Assume constant temperature on both sides .

$ p_1(t) = \frac{p_0}{2}(1- e^{-\frac{A v}{2V} t} )$

$ p_1(t) = \frac{p_0}{2}(1+ e^{-\frac{A v}{2V} t} )$

$ p_1(t) = \frac{p_0}{2}(1+ e^{\frac{A v}{2V} t} )$

$ p_1(t) = \frac{p_0}{2}(1- e^{\frac{A v}{2V} t} )$