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## Homework Statement

1 kg air at the pressure ##10^6##Pa and the temperature ##125^\circ C = 398K## expand until the volume is 5 times larger. The expansion is done with change in heat at every moment being ##1/4## of the work done by the gas. Calculate the end pressure.

## Homework Equations

##dU = dQ+dW##

##dU = nC_vdT##

##W = -\int pdV##

## The Attempt at a Solution

We have that ##dQ + dW = \frac{5}{4}dW## and the two first equations can then be combined into

## \frac{5}{4}dW = nC_vdT##

And if ##dW = -pdV## from the third equation we get

## nC_vdT = -\frac{5}{4}pdV##

Assuming an ideal gas and for air ##C_v = \frac{5}{2}R##

## nC_vdT = -\frac{5}{4}nRT\frac{dV}{V} \Longleftrightarrow 2\int \frac{dT}{T} = -\int \frac{dV}{V}

\Longleftrightarrow ln(\frac{T_2}{T_1}) = -\frac{1}{2} ln(\frac{V_2}{V_1})##

Using that ##\frac{V_2}{V_1} = 5## we get

##T_2 = T_1\frac{1}{\sqrt{5}} \approx 178K##

Using the ideal gas law again to calculate the starting volume with ##n = 1000/29 \approx 35.5 mol##

##V_1 = \frac{nRT_1}{p_1} \approx 0.114 m^3##

##V_2 = 5V_1 \approx 0.57 m^3##

And invoking the ideal gas law a third time

##P_2 = \frac{nRT_2}{V_2} \approx 92 kPa##

The correct answer according to the book should be ##123kPa##. Any ideas where i go wrong?