Question
A balloon of volume \(V\) contains \(10 \mathrm{mg}\) of an ideal gas at a pressure \(P\). An additional mass of the gas is added without changing the temperature of the balloon. This change causes the volume to increase to \(2 V\) and the pressure to increase to \(3 P\).
What is the mass of gas added to the balloon?
A. \(5 \mathrm{mg}\)
B. \(15 \mathrm{mg}\)
C. \(50 \mathrm{mg}\)
D. \(60 \mathrm{mg}\)
▶️Answer/Explanation
Ans:C
1st situation
\(P V=n RT\)
\[
P V=\left(\frac{10}{m}\right) R \cdot T…….(1)
\]
2nd situation
\[
3 P \times 2 V=\left(\frac{10+m}{m}\right) R T \text {……(2) }
\]
Dividing 2 by 1,
$6=\frac{10+m}{10}\Rightarrow m=50~mg$
Question
A fixed mass of an ideal gas expands slowly at constant temperature in a container.
Three statements about the gas molecules during the expansion are:
I. They collide with the walls of the container at a reduced rate.
II. They travel further on average between each collision.
III. Their average kinetic energy decreases as the gas expands.
Which statements are correct?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
▶️Answer/Explanation
Ans:A
During a slow, constant-temperature expansion of an ideal gas, several changes occur in the behavior of gas molecules. Let’s analyze each statement:
I. They collide with the walls of the container at a reduced rate.
This statement is correct. During a slow expansion at constant temperature, the volume of the gas increases, which means there is more space for gas molecules to occupy. As a result, the density of gas molecules decreases, and they collide with the walls of the container at a reduced rate. This is consistent with the ideal gas law, which states that at constant temperature and pressure, the volume and density of the gas are inversely proportional.
II. They travel further on average between each collision.
This statement is correct. As the volume of the gas increases during the expansion, the average distance between gas molecules (mean free path) also increases. This means that gas molecules, on average, travel further between each collision. This is a characteristic of slow, constant-temperature expansion in ideal gases.
III. Their average kinetic energy decreases as the gas expands.
This statement is incorrect. During a slow, constant-temperature expansion, the temperature of the gas remains constant. According to the ideal gas law, the product of pressure and volume is proportional to the temperature (PV = nRT). If the temperature is constant, then the product of pressure and volume must also remain constant. This implies that the average kinetic energy of gas molecules remains constant. The individual kinetic energies of gas molecules may change, but on average, they do not decrease.
So, the correct statements are I and II.
Question
Two containers X and Y are maintained at the same temperature. X has volume 4 m3 and Y has volume 6 m3.
They both hold an ideal gas. The pressure in X is 100 Pa and the pressure in Y is 50 Pa. The containers are then joined by a tube of negligible volume. What is the final pressure in the containers?
A 70 Pa
B 75 Pa
C 80 Pa
D 150 Pa
Answer/Explanation
Answer – A
\(Px.Vx + Py.Vy = Pf.Vf\)
\(100×4 + 50×6 = pf . 10\)
\(Pf= 70 pa\)
Q and R are two rigid containers of volume 3V and V respectively containing molecules of the same ideal gas initially at the same temperature. The gas pressures in Q and R are p and 3p respectively. The containers are connected through a valve of negligible volume that is initially closed.
The valve is opened in such a way that the temperature of the gases does not change. What is the change of pressure in Q?
A. +p
B. \(\frac{{ + p}}{2}\)
C. \(\frac{{ – p}}{2}\)
D. –p
Answer/Explanation
Answer – B
\( Pi_{q}Vi_{q}+ Pi_{r}Vi_{r}=P_{f}.V_{f}\)
\(P(3V)+3P(V)=Pf.4V\)
\(Pf=1.5P\)
\(\Delta P=1.5P-P\)
\(\Delta P=0.5P\)
Two containers, X and Y, are each filled by an ideal gas at the same temperature. The volume of Y is half the volume of X. The number of moles of gas in Y is three times the number of moles of the gas in X. The pressure of the gas in X is PX and the pressure of the gas in Y is PY.
What is the ratio \(\frac{{{P_X}}}{{{P_Y}}}\)?
A. \(\frac{1}{6}\)
B. \(\frac{2}{3}\)
C. \(\frac{3}{2}\)
D. 6
Answer/Explanation
Answer – A
\( \frac{P_{x}.V_{x}}{P_{y}.V_{y}}=\frac{N_{x}RT_{x}}{N_{y}RT_{y}}\)
\(V_{y}=\frac{1}{2}.V_{y}\)
\(N_{y}=3N_{x}\)
\(\frac{P_{x}}{P_{y}}.2=\frac{1}{3}\)
\(\frac{P_{x}}{P_{y}}=\frac{1}{2}\)