# What is the Maximum Temperature for a Copper Block at 1000 atm Pressure?

• Muthumanimaran
In summary, the problem involves a block of copper at a pressure of 1 atm and a temperature of 5 °C, which is then raised to 10 °C at constant volume. The final pressure is found to be 40200 Pa using the formula (Pf-Pi)=(volume expansivity / isothermal compressibility) (Tf-Ti). For part (b), the highest temperature that the system can be raised to is 17 °C, despite getting a temperature of 2.4 °C for a pressure of 1000 atm.
Muthumanimaran

## Homework Statement

(a) A block of copper at a pressure of 1 atm and a temperature of 5 degree celcuis is kept at constant volume. If the temperature is raised to 10 degree celcius what will be the final pressure?

(b) If the vessel holding the block of copper has a negligibly small thermal expansivity and can withstand a maximum pressure of 1000atm, what is the highest temperature to which the system may be raised?

(Note: The volume expansivity and isothermal compressiblity are not always listed in handbooks of data. However, volume expansivity is three times the linear expansion coefficient and isothermal compressibility is reciprocal of the bulk modulus. For this problem, assume that volume expansivity and isothermal compressibility remain practically constant within the temperature range of 0 to 20 degree celcius at the values of 4.95 x 10^(-5) K^-1 and 6.17 x 10^(-12))

## Homework Equations

(Pf-Pi)=(volume expansivity / isothermal compressibility) (Tf-Ti)

## The Attempt at a Solution

part (a) of the problem can be easily done by using by substituting the given data in the formula above.
$$P_{f}-P_{i}=\frac{\beta}{\kappa}(T_{f}-T_{i})$$
$$P_{f}-10^{5}=\frac{4.95 \times 10^{-5}}{6.17 \times 10^{-12}} (10-5)$$
gives Pf=40200 Pa

If I do the same for part (b) by taking initial temperature as 0 degree celcius and initial pressure as 1 atm, the final temperature is 2.4 degree celcius but the correct answer is 17 celcius, give me a hint.

Last edited:

DrClaude said:
$$P_{f}-P_{i}=\frac{\beta}{\kappa}(T_{f}-T_{i})$$
$$P_{f}-10^{5}=\frac{4.95 \times 10^{-5}}{6.17 \times 10^{-12}} (10-5)$$
gives Pf=40200 Pa

Muthumanimaran said:
Pf=40200 Pa
You are off by a few orders of magnitude.

In any case, your result for Tf = 10 °C is Pf < 1000 atm. How can you get a temperature of 2.4 °C for 1000 atm?

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