MHT CET · Chemistry · States of Matter
What is the volume of a gas at \(1.032 \times 10^5 \mathrm{Nm}^{-2}\) if it occupies \(1 \mathrm{dm}^3\) of volume at normal temperature and pressure?
- A \(0.982 \cdot \mathrm{dm}^3\)
- B \(1.3 \mathrm{dm}^3\)
- C \(1.5 \mathrm{dm}^3\)
- D \(1.7 \mathrm{dm}^3\)
Answer & Solution
Correct Answer
(A) \(0.982 \cdot \mathrm{dm}^3\)
Step-by-step Solution
Detailed explanation
\(1 \mathrm{Nm}^{-2}=1 \mathrm{~Pa}=0.9869 \times 10^{-5} \mathrm{~atm}\)
Normal pressure \(=1 \mathrm{~atm}\)
\(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2\)
\(\mathrm{V}_1=\frac{1 \mathrm{~atm} \times 1 \mathrm{dm}^3}{\left(1.032 \times 10^5 \times 0.987 \times 10^{-5}\right) \mathrm{atm}}\) \(=0.982 \mathrm{dm}^3\)
Normal pressure \(=1 \mathrm{~atm}\)
\(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2\)
\(\mathrm{V}_1=\frac{1 \mathrm{~atm} \times 1 \mathrm{dm}^3}{\left(1.032 \times 10^5 \times 0.987 \times 10^{-5}\right) \mathrm{atm}}\) \(=0.982 \mathrm{dm}^3\)
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