Search any question & find its solution
Question:
Answered & Verified by Expert
A gas volume \(251 \mathrm{~cm}^3\) at \(20^{\circ} \mathrm{C}\) and pressure \(78 \mathrm{~cm}\) of \(\mathrm{Hg}\). Find its volume at NTP
Options:
Solution:
2175 Upvotes
Verified Answer
The correct answer is:
\(263.8 \mathrm{~cm}^3\)
Volume of gas, \(V=251 \mathrm{~cm}^3\)
Temperature,
\(\begin{aligned}
& T_1=20^{\circ} \mathrm{C}=273+20=293^{\circ} \mathrm{K} \\
& p_1=78 \mathrm{~cm} \text { of } \mathrm{Hg}
\end{aligned}\)
At NTP,
Temperature, \(T_2=300 \mathrm{~K}\)
Pressure,
\(\begin{aligned}
& p_2=76 \mathrm{~cm} \text { of } \mathrm{Hg} \\
& V_2=?
\end{aligned}\)
According to ideal gas equation,
\(\begin{aligned}
& \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \\
\Rightarrow & V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{78 \times 251 \times 300}{76 \times 293} \\
& =263.8 \mathrm{~cm}^3
\end{aligned}\)
No option is matched.
Temperature,
\(\begin{aligned}
& T_1=20^{\circ} \mathrm{C}=273+20=293^{\circ} \mathrm{K} \\
& p_1=78 \mathrm{~cm} \text { of } \mathrm{Hg}
\end{aligned}\)
At NTP,
Temperature, \(T_2=300 \mathrm{~K}\)
Pressure,
\(\begin{aligned}
& p_2=76 \mathrm{~cm} \text { of } \mathrm{Hg} \\
& V_2=?
\end{aligned}\)
According to ideal gas equation,
\(\begin{aligned}
& \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \\
\Rightarrow & V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{78 \times 251 \times 300}{76 \times 293} \\
& =263.8 \mathrm{~cm}^3
\end{aligned}\)
No option is matched.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.