find the mean proportion between 1.8and 0.2
Answers
Step-by-step explanation:
Since the temperature of the system remains constant, Boyle's Law is applicable here, i.e., the product of pressure and volume remains constant.
\sf{\longrightarrow P_iV_i=P_fV_f}⟶P
i
V
i
=P
f
V
f
Let the volume of the large container be V, then volume of small container will be V/3.
Pressure of gas in large container is 10.3 atm and that of gas in small container is 1.1 atm.
Initial product of pressure and volume of the system,
\longrightarrow\sf{P_i}\sf{V_i}=\sf{10.3V}+\sf{\dfrac{1.1V}{3}}⟶P
i
V
i
=10.3V+
3
1.1V
When the two cylinders are connected, the volume of the system becomes V + V/3 = 4V/3. Let P be the final pressure of the system.
Final product of pressure and volume of the system,
\longrightarrow\sf{P_fV_f}=\sf{\dfrac{4PV}{3}}⟶P
f
V
f
=
3
4PV
Now,
\longrightarrow\sf{10.3V}+\sf{\dfrac{1.1V}{3}}=\sf{\dfrac{4PV}{3}}⟶10.3V+
3
1.1V
=
3
4PV
\longrightarrow\sf{\dfrac{4P}{3}}=\sf{10.3}+\sf{\dfrac{1.1}{3}}⟶
3
4P
=10.3+
3
1.1
\longrightarrow\sf{\dfrac{4P}{3}}=\sf{\dfrac{32.0}{3}}⟶
3
4P
=
3
32.0
\longrightarrow\underline{\underline{\sf{P}=\sf{8.0\ atm}}}⟶
P=8.0 atm