KCET · Chemistry · States of Matter
When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas
- A will be half of original volume
- B will be 4 times the original volume
- C will be 2 times the original volume
- D will be \(1 / 4\) th times the original volume
Answer & Solution
Correct Answer
(B) will be 4 times the original volume
Step-by-step Solution
Detailed explanation
Given, final temperature \(\left(T_{2}\right)\) \(=2 \times\left(T_{1}\right)\) initial temperatue Final pressure \(\left(p_{2}\right)=2 \times \frac{\left(p_{1}\right)}{2}\) initial pressure
According to gas law,
\(\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}\)
\(\therefore \quad \frac{p_{1} V_{1}}{T_{1}}=\frac{p_{1} \times V_{2}}{2 \times 2 T_{1}}\)
\(V_{2}=4 V_{1}\)
\(\therefore\) Volume increase by 4 times.
According to gas law,
\(\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}\)
\(\therefore \quad \frac{p_{1} V_{1}}{T_{1}}=\frac{p_{1} \times V_{2}}{2 \times 2 T_{1}}\)
\(V_{2}=4 V_{1}\)
\(\therefore\) Volume increase by 4 times.
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