Question

Please someone help me to solve this problem​

Answer 1

(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.

Step-by-step explanation:

(a)

`P_(2)=(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)=(1)/(2)T_(1)`

`P_(2)=(P_(1)* V_(1)* (1)/(2)T_(1))/(T_(1)*2V_(1))=(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)=(P_(1)* V_(1)* 2T_(1))/(T_(1)*2V_(1))={P_(1)}`

Therefore, the pressure of the sample unchanged.

(c)

Volume of sample halved and temperature double.

`V_(2)=(1)/(2)V_(1)`

`T_(2)=2T_(1)`

`P_(2)=(P_(1)* V_(1)* 2T_(1))/(T_(1)* (1)/(2)V_(1))={4P_(1)}`

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