Target: 10 Questions in 10 minutes

An IB Chemistry data booklet is helpful

1. Which of the following is NOT a characteristic of an ideal gas?
  • A.   There is a weak attractive force between the molecules
  • B.   Gas molecules occupy a negligible volume
  • C.   Gas molecules are in random motion
  • D.   Only elastic collisions occur between molecules
2. Which graph represents the relationship between the pressure (in Pa) and the volume (in dm3) of a fixed mass of an ideal gas at constant temperature?

P ~ V graphs for gases x 4

3. The deviation between real gas and ideal gas behavior is greatest at ...
  • A.   high temperature and low pressure
  • B.   high temperature and high pressure
  • C.   low temperature and low pressure
  • D.   low temperature and high pressure
4. What will happen to the pressure of a fixed mass of gas if the temperature (in K) and the volume are both doubled?
  • A.   It will remain the same
  • B.   It will increase by a factor of 2
  • C.   It will increase by a factor of 4
  • D.   It will decrease

5. A fixed mass of a gas occupies 40cm3 at 26.85°C. At what temperature, in °C, will the volume of gas be 80cm3 if the pressure remains constant?

  • A.   53.85
  • B.   149.85
  • C.   326.85
  • D.   599.85
6. The pressure and temperature (in K) of a fixed mass of gas are both halved. If the initial volume of the gas was 100cm3, what is the new volume?
  • A.   50cm3
  • B.   100cm3
  • C.   200cm3
  • D.   400cm3

7. Which graph represents the relationship between the volume (in cm3) and the temperature (in °C) of a fixed mass of an ideal gas at constant pressure?

V~T graph for a gas x 4

8. How many moles of gas are in a container of volume 2000cm3 at 100kPa pressure and a temperature of 400K?
(R = 8.31 J K-1 mol-1)
 
  • A. 
$${100 \times 2000} \over{8.31 \times 400}$$
  • B. 
$${100 \times 2} \over{8.31 \times 400}$$
  • C. 
$${100 \times 2000} \over{8.31 \times 127}$$
  • D. 
$${100 000 \times 2000} \over{8.31 \times 400}$$
9. A fixed mass of gas is at a pressure of 4kPa.

The volume of the gas is halved and the temperature (in K) of the gas is doubled.
What is the new gas pressure after these changes?

  • A.   1kPa
  • B.   4kPa
  • C.   8kPa
  • D.   16kPa
10. What is the molecular mass of gas Z if 2.00g of the gas occupies a volume of 500cm3 at 24.85°C and 1.01 x 105 Pa?
(R = 8.31 J K-1 mol-1)
 
  • A. 
$${2.00 \times 8.31 \times 24.85} \over{1.01 \times 10^5 \times 0.5}$$
  • B. 
$${1.01 \times 10^5 \times 0.5} \over{2.00 \times 8.31 \times 298}$$
  • C. 
$${2.00 \times 8.31 \times 298} \over{1.01 \times 10^5 \times 5 \times 10^{-4}}$$
  • D. 
$${1.01 \times 10^5 \times 5 \times 10^{-4}} \over{2.00 \times 8.31 \times 298}$$
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Question 1:

Let’s go through each option:

A. There is a weak attractive force between the molecules
In an ideal gas, it is assumed that there are no intermolecular forces. So weak attractive forces are not a characteristic of an ideal gas — this statement is false for an ideal gas.

B. Gas molecules occupy a negligible volume
This is true for an ideal gas (point particles).

C. Gas molecules are in random motion
This is true for an ideal gas (kinetic theory).

D. Only elastic collisions occur between molecules
This is true for an ideal gas (kinetic energy conserved in collisions).

Thus, the one that is NOT a characteristic is A.


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Question 2:

The relationship between the pressure and volume of a fixed mass of an ideal gas at a constant temperature is described by Boyle's Law.

According to Boyle's Law, pressure is inversely proportional to volume. This means as the volume of the gas increases, its pressure decreases, and vice versa.

The graph that best represents this relationship is a downward-sloping curve (a hyperbola) when pressure (P) is plotted against volume (V). The equation for this relationship is PV=k, where k is a constant.

This is shown by graph A.


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Question 3:

For an ideal gas, we assume:

Real gases deviate from ideal behavior when:

  1. Pressure is high → molecules are forced closer together, so the volume of molecules is no longer negligible.

  2. Temperature is low → molecules have less kinetic energy, so intermolecular attractions become significant.

Thus, the greatest deviation occurs at low temperature and high pressure.

Answer D

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Question 4:

We use the ideal gas law:

PV=nRT

Here, n and R are constant.

Initially: P1V1=nRT1

After changes:

New equation:

P2(2V1)=nR(2T1)

But nRT1=P1V1, so:

P2(2V1)=2(P1V1)

Divide both sides by 2V1:

P2=P1

So the pressure remains the same. This corresponds to answer A.

 


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Question 5:

We can solve this using Charles’s law:

V1/T1=V2/T2

where temperature is in kelvin.


Step 1: Convert given temperature to Kelvin

T1=26.85∘C+273.15=300.00 K
V1=40 cm3, V2=80 cm3

Step 2: Solve for T2:

40/300=80/T2

So T2=80 × 300/40 = 80 × 7.5 = 600 K


Step 3: Convert T2 back to °C

T2(in ∘C)=600−273.15=326.85∘C

Step 4: Match with options

C: That corresponds to 326.85°C.

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Question 6:

We can use the combined gas law:

P1V1/T1=P2V2/T2

Step 1: Identify given values


Step 2: Substitute into the equation

P1⋅100/T1= ½P1⋅V2 /½T1

Step 3: Simplify the right-hand side

P1⋅100/T1= (½/½)P1⋅V2 /T1 = P1⋅V2 /T1


Step 4: Equate both sides

Cancel P1/T1from both sides:

100=V2

Step 5: Conclusion

The new volume is 100 cm3. Answer B.


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Question 7:

The graph that represents the relationship between the volume and the temperature in kelvin of a fixed mass of an ideal gas at constant pressure is a straight line passing through the origin of the Kelvin temperature scale.

This relationship is described by Charles's Law.

Charles's Law Explained
Charles's Law states that for a fixed mass of an ideal gas at constant pressure, the volume (V) is directly proportional to its absolute temperature (T).

Mathematically:

V∝T or V = kT

=k(where k is a constant)
If you plot Volume (y-axis) against Temperature (x-axis):

Using the Kelvin Scale (Absolute Temperature): The graph is a straight line that passes through the origin (0,0), demonstrating the direct proportionality. At 0 K, the theoretical volume of the gas is zero.

Using the Celsius Scale: The graph is still a straight line, but it does not pass through the origin. Instead, the line extrapolates (extends backward) to the x-axis (zero volume) at a temperature of −273.15 °C (Absolute Zero).

Since the question asks for the temperature in °C, the correct graph is a straight line that, when extrapolated, intersects the x-axis at −273.15 °C.

*The line will therefore be straight, but have a positive value at 0°C. This corresponds to graph C.
The Silverback (edit )


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Question 8:

We use the ideal gas law:

PV=nRT

Step 1: Convert units to match R (8.31 J K⁻¹ mol⁻¹)


Step 2: Solve for n

n=PV/RT

Substitute:

n= (1.0×105)×(2.0×10−3)/ (8.31×400)

Step 3: Match with given options

B: 100 x 2 is the same as our (1.0×105)×(2.0×10−3)

That’s correct.

* The Silverback (edit - removed the other options which are incorrect, for brevity)


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Question 9:

We use the combined gas law:

P1V1/T1=P2V2/T2

Step 1: Identify given values


Step 2: Substitute into the equation

4⋅V1/T1=P2⋅½V1/2T1

Step 3: Simplify the right-hand side

P2⋅½V1/2T1= P2⋅V1/4T1

So:

4⋅V1/T1=P2⋅V1/4T1

Step 4: Cancel V1/T1from both sides

4=P2 / 4, so P2 = 16 (kPa)

Step 5: Conclusion

The new pressure is 16 kPa

Answer D

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Question 10:

We use the ideal gas law:

PV=nRT

where n=m/M, with m= mass, M = molar mass.

So:

PV=(m/M)RT

Step 1: Identify given values in correct units


Step 2: Write expression for M

M=mRT / PV

M=(2.00×8.31×298) / (1.01×105×5.00×10-4)

Step 3: Match with options

That matches C exactly:

M=(2.00×8.31×298) / (1.01×105×5.00×10-4)

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