Question

1.A container holds 0.320 m^3 of oxygen at an absolute pressure of 6.00 atm. A value is opened, allowing 1/3 the gas to escape. If the volume of the container and the temperature remain constant through1 the process, what is the final pressure of the gas?

2.An 18.00 cm^3 cube suspended from a scale has a weight of 1.99 N as shown in the left picture. When the cube is submerged in water as in the right picture, what will the scale read?

Answer 1

1. Given:

V1 = initial volume =

P1 = initial pressure = 6 atm

T1 = initial temperature

n1 = initial number of moles

When the amount of gas escaped 1/3 of original, then

V2 = final volume = V1

T2 = final temperature = T1

n2 = final number of moles = $n_{1}-\frac{1}{3}n_{1}=\frac{2}{3}n_{1}$

P2 = final temperature = ?

The ideal gas equation is:

$PV=nRT$

-----(i)

RT 21 n V

As the volume and temperature is kept constant, then using the equation (i):

$\frac{P_{2}}{n_{2}}=\frac{P_{1}}{n_{1}}$

$\Rightarrow P_{2}=\frac{P_{1}}{n_{1}}*n_{2}$

$\Rightarrow P_{2}=\frac{6}{n_{1}}*\frac{2}{3}n_{1} = \frac{6*2}{3}=4 atm$

Therefore, the final pressure of the gas is 4.00 atm. [Answer]