Question

(13%) Problem 4: One mole of an ideal gas does -1700 J of work as it expands isothermally to a final pressure of 1.00 atm and volume of 0.029 m³. 50% Part (a) What was the initial volume of the gas, in cubic meters? V1=7.66 * 10^(3) 50% Part (b) What is the temperature of the gas, in kelvin? T=353.43 T=353.4 \checkmark Correct!

          (13%) Problem 4: One mole of an ideal gas does -1700 J of work as it expands isothermally to a final pressure of 1.00 atm and volume of 0.029 m³.
50% Part (a) What was the initial volume of the gas, in cubic meters?
V1=7.66 * 10^(3)
50% Part (b) What is the temperature of the gas, in kelvin?
T=353.43
T=353.4 \checkmark Correct!

(13%) Problem 4: One mole of an ideal gas does -1700 J of work as it expands isothermally to a final pressure of 1.00 atm and volume of 0.029 m³.
50% Part (a) What was the initial volume of the gas, in cubic meters?
V1=7.66 * 10^(3)
50% Part (b) What is the temperature of the gas, in kelvin?
T=353.43
T=353.4 Correct!

Added by Michael A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

11/08/2022

Step 1

Given that one mole of an ideal gas does -1700 J of work as it expands isothermally, we can use the formula for work done during an isothermal process: W = -nRT * ln(V_2/V_1) where W is the work done (-1700 J), n is the number of moles (1 mole), R is the gas Show more…

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13% Problem 4: One mole of an ideal gas does -1700 J of work as it expands isothermally to a final pressure of 1.00 atm and volume of 0.029 m^3. 50% Part (a) What was the initial volume of the gas, in cubic meters? V_1 = 7.66 × 10^-3 m^3 50% Part (b) What is the temperature of the gas, in kelvin? T = 353.4 K Note: The text contains a mix of correct and incorrect formatting, and some symbols are not properly rendered. The corrected text above assumes that the initial volume is 7.66 × 10^-3 m^3 and the temperature is 353.4 K, as indicated by the numbers provided. The square root symbol is not present in the text, so no correction was needed for that.