Solveeit Logo

Question

Question: At constant temperature, if pressure increases by \(1\% \), the percentage decrease of volume is: ...

At constant temperature, if pressure increases by 1%1\% , the percentage decrease of volume is:
A) 1%1\%
B) 100101%\dfrac{{100}}{{101}}\%
C) 1100%\dfrac{1}{{100}}\%
D) 1101%\dfrac{1}{{101}}\%

Explanation

Solution

To solve this we must know the law that gives the relationship between volume and pressure at constant temperature. Boyle's law gives the relationship between volume and pressure at constant temperature. Consider the initial pressure to be 100 mm100{\text{ mm}} and then solve.

Complete solution:
We know that Boyle’s law states that at constant temperature the volume of an ideal gas is inversely proportional to the pressure of the gas. Thus,
P1V1=P2V2{P_1}{V_1} = {P_2}{V_2} …… (1)
Where P1{P_1} is the initial pressure,
P2{P_2} is the final pressure,
V1{V_1} is the initial volume,
V2{V_2} is the final volume.
Assume that the initial pressure is 100 mm100{\text{ mm}}. Thus,
P1=100 mm{P_1} = 100{\text{ mm}}
We are given that the pressure increases by 1%1\% . Thus, the final pressure will be,
P2=101 mm{P_2} = 101{\text{ mm}}
Substitute the values for the initial pressure and final pressure in equation (1). Thus,
100 mm×V1=101 mm×V2100{\text{ mm}} \times {V_1} = 101{\text{ mm}} \times {V_2}
V2=100 mm101 mm×V1{V_2} = \dfrac{{100{\text{ mm}}}}{{101{\text{ mm}}}} \times {V_1}
The decrease in the volume will be,
Decrease in volume =V1100 mm101 mm×V1 = {V_1} - \dfrac{{100{\text{ mm}}}}{{101{\text{ mm}}}} \times {V_1}
Decrease in volume =100101%= \dfrac{{100}}{{101}}\%
Thus, the percentage decrease of volume is 100101%\dfrac{{100}}{{101}}\% .
Thus, the correct option is (B) all of the above.

Note: We know that Boyle's law states that at constant temperature the volume of an ideal gas is inversely proportional to the pressure of the gas. Thus, as the volume of the gas increases, the pressure decreases or as the volume of the gas decreases, the pressure increases.