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5.12     Calculate the temperature of 4.0 \: mol of a gas occupying 5\: \: dm^{3} at 3.32 bar.
                 (R = 0.083 \: bar\: dm^{3} K^{-1} mol^{-1}).

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We have the ideal gas equation,

PV=nRT 

Pressure, p=3.32bar

Volume, V=5 dm^3

Number of moles, n=4 mol

Gas constant R= 0.083

Temperature,  T=?

T = \frac{PV}{nR} = \frac{3.32bar\times 5\ dm^3}{4\ mol\times 0.083 bar\ dm^3\ K^{-1}mol^{-1}} = 50K

so, temperature, T=50K

Posted by

Divya Prakash Singh

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