Answer:
In the rigid container : pressure increases when temperature increases and density remains constant
In the flexible container: pressure remains constant and density decreases when temperature increases
Step-by-step explanation:
Using the Ideal gas law
![P*V=n*R*T](https://img.qammunity.org/2020/formulas/physics/middle-school/qdt0l7q9iqq9ychn9hfug6ate3urcsv6zo.png)
P= absolute pressure, V=volume , n= number of moles (mass) , R=constant , T= absolute temperature
And
![D= m/V](https://img.qammunity.org/2020/formulas/chemistry/high-school/myts6auz1wrci30jwczylslbfh44gl4kv5.png)
D= Density , m = mass , V= volume
In both containers, since they are sealed, the mass is kept inside and remains constant --> n= constant and m= constant
- In the rigid container , V = constant , therefore
![P*V=n*R*T\\\\P=n*R*T/V = (n*R/V) * T = constant * T\\\\P = constant * T](https://img.qammunity.org/2020/formulas/chemistry/high-school/tpzfxgkkazvngduglutrsyiu2r39qfca5c.png)
And thus absolute pressure increases with absolute temperature
Regarding density
![D= m/V = constant 1 /constant 2= constant 3](https://img.qammunity.org/2020/formulas/chemistry/high-school/28yt8xidullg9llj6uw6i71kl5nxxz4vad.png)
Thus density remains constant
- In the flexible container , P = constant , therefore
![P*V=n*R*T\\\\V=n*R*T/P = (n*R/P) * T = constant * T\\\\V = constant * T](https://img.qammunity.org/2020/formulas/chemistry/high-school/ceguo7mf6hlvg3k8xxer8p63nadevo962e.png)
but also
![D= m/V = constant 1 /(constant 2* T)= constant 3 / T\\\\D = constant 3* T](https://img.qammunity.org/2020/formulas/chemistry/high-school/9hfonookh9kkqkc8l565ytj2fmh958kxz0.png)
And thus density decreases with temperature