Question

Please someone help me to solve this problem​

Answer 1

Answer:

(a) The final pressure of the sample becomes one-fourth of the original pressure.

(b) The pressure of the sample remains unchanged.

(c) The final pressure of the sample becomes four times of the original pressure.

Explanation:

(a)

$P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$

The volume of sample doubled and kelvin temperature halved.

$V_{2}=2V_{1}$

$T_{2}=\frac{1}{2}T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times \frac{1}{2}T_{1}}{T_{1}\times2V_{1}}=\frac{P_{1}}{4}$

Therefore, the final pressure of the sample becomes one-fourth of the original pressure.

(b)

Volume and temperature of sample doubled.

$V_{2}=2V_{1}$

$T_{2}=2T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times2V_{1}}={P_{1}}$

Therefore, the pressure of the sample unchanged.

(c)

Volume of sample halved and temperature double.

$V_{2}=\frac{1}{2}V_{1}$

$T_{2}=2T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times \frac{1}{2}V_{1}}={4P_{1}}$

Therefore, the pressure of the sample becomes four times of the original pressure.