KCET · Physics · Kinetic Theory of Gases
A perfect gas at \(27^{\circ} \mathrm{C}\) is heated at constant pressure so as to double its volume. The increase in temperature of the gas will be
- A \(600^{\circ} \mathrm{C}\)
- B \(327^{\circ} \mathrm{C}\)
- C \(54^{\circ} \mathrm{C}\)
- D \(300^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(300^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
We know for perfect gas
Here, \(\quad \frac{V}{V_{2}}=\frac{T_{1}}{T_{2}}\)
According to question,
\(V_{1}=V\) then \(V_{2}=2 V\) and \(T_{1}=300 \mathrm{~K}\)
\(\therefore \frac{1}{2} =\frac{300}{T_{2}} \)
\(T_{2} =600 \mathrm{~K} \)
\(T_{2} =327^{\circ} \mathrm{C}\)
So, \(\Delta t=327-27=300^{\circ} \mathrm{C}\)
Here, \(\quad \frac{V}{V_{2}}=\frac{T_{1}}{T_{2}}\)
According to question,
\(V_{1}=V\) then \(V_{2}=2 V\) and \(T_{1}=300 \mathrm{~K}\)
\(\therefore \frac{1}{2} =\frac{300}{T_{2}} \)
\(T_{2} =600 \mathrm{~K} \)
\(T_{2} =327^{\circ} \mathrm{C}\)
So, \(\Delta t=327-27=300^{\circ} \mathrm{C}\)
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