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Consider Cu(s) at 293 K. Its isothermal compressibility is 7.35 × 10^-7 atm^-1. What pressure needs to be applied to increase its density by 0.08%?

a. 7.35 atm
b. 73.5 atm
c. 735 atm
d. 7350 atm

User Naoto Ida
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1 Answer

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Final answer:

To calculate the pressure needed to increase the density of Cu(s) by 0.08%, we can use the equation: ΔP = P * β * ΔV where ΔP is the change in pressure, P is the initial pressure, β is the isothermal compressibility, and ΔV is the change in volume.

Given that the isothermal compressibility of Cu(s) is 7.35 × 10^-7 atm^-1 and the change in density is 0.08%, we can calculate: ΔP = 1 * 7.35 × 10^-7 * 0.0008 = 5.88 × 10^-10 atm. Therefore, the pressure needed to increase the density of Cu(s) by 0.08% is approximately 5.88 × 10^-10 atm.

Step-by-step explanation:

To calculate the pressure needed to increase the density of Cu(s) by 0.08%, we can use the equation:

ΔP = P * β * ΔV

where ΔP is the change in pressure, P is the initial pressure, β is the isothermal compressibility, and ΔV is the change in volume. we can calculate: ΔP = 1 * 7.35 × 10^-7 * 0.0008 = 5.88 × 10^-10 atm. Therefore, the pressure needed to increase the density of Cu(s) by 0.08% is approximately 5.88 × 10^-10 atm.

Given that the isothermal compressibility of Cu(s) is 7.35 × 10^-7 atm^-1 and the change in density is 0.08%, we can calculate:

ΔP = 1 * 7.35 × 10^-7 * 0.0008 = 5.88 × 10^-10 atm

Therefore, the pressure needed to increase the density of Cu(s) by 0.08% is approximately 5.88 × 10^-10 atm.

User Roman Pushkin
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7.1k points