AS & A Level Chemistry 4.1 The gaseous state , ideal and real gases and $p V=n R T$ Exam Style Practice Questions Paper 1- New Syllabus
Question
A pure sample of a gas has a density of \(2.62\,\mathrm{g\,dm^{-3}}\) at \(101000\,\mathrm{Pa}\) and \(25^\circ\mathrm{C}\). The gas behaves ideally under these conditions.
Which expression gives the \(M_r\) of the gas?
(B) \(\dfrac{101000\times0.001}{2.62\times8.31\times298}\)
(C) \(\dfrac{2.62\times8.31\times25}{101000\times0.001}\)
(D) \(\dfrac{2.62\times8.31\times298}{101000\times0.001}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
For an ideal gas:
\(PV=nRT\)
Using \(n=\dfrac{m}{M_r}\) and density \(=\dfrac{m}{V}\):
\(M_r=\dfrac{\mathrm{density}\times RT}{P}\)
Temperature must be converted to kelvin: \(25^\circ\mathrm{C}=298\,\mathrm{K}\), and \(1\,\mathrm{dm^3}=0.001\,\mathrm{m^3}\).
Therefore, the correct expression is (D).
Question
\(1.00\,\mathrm{g}\) of nitrogen gas is stored in a \(2.00\,\mathrm{dm^3}\) vessel at \(40.0^\circ\mathrm{C}\).
What is the pressure in the vessel?
(B) \(11\,900\,\mathrm{Pa}\)
(C) \(46\,400\,\mathrm{Pa}\)
(D) \(92\,900\,\mathrm{Pa}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Moles of nitrogen:
\(n=\dfrac{1.00}{28.0}=0.0357\,\mathrm{mol}\)
Convert the volume and temperature:
\(V=2.00\,\mathrm{dm^3}=2.00\times10^{-3}\,\mathrm{m^3}\)
\(T=40.0+273=313\,\mathrm{K}\)
Using the ideal gas equation, \(PV=nRT\).
\(P=\dfrac{nRT}{V}=\dfrac{0.0357\times8.31\times313}{2.00\times10^{-3}}\approx4.64\times10^4\,\mathrm{Pa}\)
Therefore, the correct answer is (C).
Question
Two glass vessels, M and N, are connected by a closed valve.
M contains helium at \(20^\circ\mathrm{C}\) at a pressure of \(1.0\times10^5\,\mathrm{Pa}\). N has been evacuated and has three times the volume of M.

The valve is opened and the temperature of the whole apparatus is raised to \(100^\circ\mathrm{C}\).
What is the final pressure in the system?
(B) \(4.24\times10^4\,\mathrm{Pa}\)
(C) \(1.25\times10^5\,\mathrm{Pa}\)
(D) \(5.09\times10^5\,\mathrm{Pa}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Initially:
\(P_1=1.0\times10^5\,\mathrm{Pa}\), \(T_1=293\,\mathrm{K}\).
The total volume after opening the valve is:
\(V_2=V+3V=4V\)
The final temperature is:
\(T_2=373\,\mathrm{K}\)
Using \( \dfrac{PV}{T}=\text{constant} \):
\(P_2=P_1\times\dfrac{V_1}{V_2}\times\dfrac{T_2}{T_1}\)
\(=1.0\times10^5\times\dfrac{1}{4}\times\dfrac{373}{293}\)
\(=3.18\times10^4\,\mathrm{Pa}\)
Therefore, the correct answer is (A).
