Question

Let the pressure P and volume V for a certain gas. If the pressure is increased by 25 percent at a constant temperature, then what would be the volume of the gas?

Answer 1

Answer:

The volume of the ideal gas is increased by 25% at a constant temperature. What is the percent decrease in pressure?

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The problem states that the gas in consideration is ideal and has undergone isothermal expansion. (As the volume has increased by 25%.)

For isothermal processes, P1∗V1=P2∗V2

P2=P1∗V1/V2=P1∗V1/1.25∗V1 (As the volume has increased by 25%, the final volume becomes V_2 = (1 + 0.25)*V_1

P2/P1=1/1.25=0.64

P2=P1∗0.64=(1−0.36)∗P1

Thus there is a decrease of 36% in pressure for 25% increase in volume for an isothermal prThe volume of the ideal gas is increased by 25% at a constant temperature. What is the percent decrease in pressure?

Are you preparing for the JEE/NEET 2022?

The problem states that the gas in consideration is ideal and has undergone isothermal expansion. (As the volume has increased by 25%.)

For isothermal processes, P1∗V1=P2∗V2

P2=P1∗V1/V2=P1∗V1/1.25∗V1 (As the volume has increased by 25%, the final volume becomes V_2 = (1 + 0.25)*V_1

P2/P1=1/1.25=0.64

P2=P1∗0.64=(1−0.36)∗P1

Thus there is a decrease of 36% in pressure for 25% increase in volume for an isothermal process.

Answer 2

Volume of gas is 0.8 times the initial volume.

Given - Pressure increased by 25%

Find - Volume of gas

Solution - Let the initial and final pressure of gas be P1 and P2 and initial and final volume of gas be V1 and V2.

P2 = P1 + 25%*P1

P2 = P1 + 0.25P1

P2 = 1.25P1

P1V1 = P2V2

P1*V1 = 1.25P1*V2

V2 = $\frac{P1*V1}{1.25P1}$

Cancelling P1 from both numerator and denominator

V2 = $\frac{V1}{1.25}$

$V2 = 0.8V1$

Hence, the final volume of gas will be 0.8 times the initial volume.

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