Question

An ideal gas is brought through an isothermal compression process. The 2.00 mol of gas goes from having an initial volume of 245.1 cm^3 to 125.0 cm^3. If 2166 cal is released by the gas during this process, what are the temperature T of the gas and the final pressure pf? The gas constant is R=8.31 J/molâK, and there are 4.19 J/cal. T= ____ K
pf = ____ Pa

Answer 1

Answer:

Temperature, T = - 537.95 K

$\rho_{f} = 1.07\times 10^{8} Pa$

Given:

no. of moles, n = 2

initial volume, V =$245.1 cm^{3} = 245.1\times 10^{- 6} m^{3}$

final volume, V = $125.0 cm^{3} = 152.0\times 10^{- 6} m^{3}$

heat, q = 2166 cal = $2166\times 4.19 = 9075.54 J$

Rydberg's constant, R = 9.31 J/mol.K

Solution:

Now, using the formula for the calculation of temperature:

$q = nRTln\frac{V'}{V}$

where

T = temperature

$9075.54 = 2\times 8.31\times T ln\frac{125.0}{245.1}$

$T = - 810.95^{\circ} = - 810.95^{\circ} + 273 = - 537.95 K$

Now, using:

$V'\rho_{f} = nRT$

$125.0\times 10^{- 6}\rho_{f} = 2\times 8.31\times - 810.95^{\circ}$

$\rho_{f} = 1.07\times 10^{8} Pa$