KCET · Chemistry · States of Matter
A gas at a pressure of \(2 \mathrm{~atm}\) is heated from \(25^{\circ} \mathrm{C}\) to \(323^{\circ} \mathrm{C}\) and simultaneously compressed of \(\frac{2}{3}\) rd of its original value. Then the final pressure is
- A \(1.33 \mathrm{~atm}\)
- B \(6 \mathrm{~atm}\)
- C \(2 \mathrm{~atm}\)
- D \(4 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(B) \(6 \mathrm{~atm}\)
Step-by-step Solution
Detailed explanation
Given, \(p_1=2 \mathrm{~atm}\)
\(\begin{aligned} p_2 & =? \\ V_1 & =1 \mathrm{~L} \\ V_2 & =\frac{3}{2} \\ T_1 & =298 \mathrm{~K} \\ T_2 & =596 \mathrm{~K}\end{aligned}\)
Using combined gas law, we get
\(\begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow \quad p_2 & =\frac{2 \times 1}{298} \times \frac{3 \times 596}{2}=6 \mathrm{~atm}\end{aligned}\)
\(\begin{aligned} p_2 & =? \\ V_1 & =1 \mathrm{~L} \\ V_2 & =\frac{3}{2} \\ T_1 & =298 \mathrm{~K} \\ T_2 & =596 \mathrm{~K}\end{aligned}\)
Using combined gas law, we get
\(\begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow \quad p_2 & =\frac{2 \times 1}{298} \times \frac{3 \times 596}{2}=6 \mathrm{~atm}\end{aligned}\)
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