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Luba_88 [7]
3 years ago
15

Please someone help me to solve this problem​

Chemistry
1 answer:
Zinaida [17]3 years ago
6 0

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.

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