MHT CET · Chemistry · States of Matter
Calculate the pressure of 1.5 mole of gas having volume \(3 \mathrm{dm}^3\) at \(300 \mathrm{~K}\left(\mathrm{R}=0.0821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(12.32 \mathrm{~atm}\)
- B \(14.6 \mathrm{~atm}\)
- C \(10.25 \mathrm{~atm}\)
- D \(15.3 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(A) \(12.32 \mathrm{~atm}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{P} \times 3=1.5 \times 0.0821 \times 300 \\ & \mathrm{P}=12.32 \mathrm{~atm} .\end{aligned}\)
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