MHT CET · Chemistry · States of Matter
3.4 moles of an ideal gas occupies volume of 68 mL at 300 K . What would be the pressure of gas? \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(1.247 \times 10^2 \mathrm{kPa}\)
- B \(2.431 \times 10^3 \mathrm{kPa}\)
- C \(1.031 \times 10^5 \mathrm{kPa}\)
- D \(3.247 \times 10^5 \mathrm{kPa}\)
Answer & Solution
Correct Answer
(A) \(1.247 \times 10^2 \mathrm{kPa}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{n}=3.4 \mathrm{~mol}, \mathrm{~T}=300 \mathrm{~K} \\
& \mathrm{~V}=68 \mathrm{~mL}=0.068 \mathrm{~L}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}\)
According to ideal gas equation,
\(\mathrm{PV}=\mathrm{nRT}\)
\(\therefore \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{~V}}\) \(=\) \(\frac{3.4 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.068 \mathrm{dm}^3}\)
\(=124710 \mathrm{kPa}\)
& \mathrm{n}=3.4 \mathrm{~mol}, \mathrm{~T}=300 \mathrm{~K} \\
& \mathrm{~V}=68 \mathrm{~mL}=0.068 \mathrm{~L}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}\)
According to ideal gas equation,
\(\mathrm{PV}=\mathrm{nRT}\)
\(\therefore \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{~V}}\) \(=\) \(\frac{3.4 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.068 \mathrm{dm}^3}\)
\(=124710 \mathrm{kPa}\)
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