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Calculate the temperature of 4.0 mole of a gas occupying \(5 \mathrm{dm}^3\) at \(3.32\) bar. \(\left(\mathrm{R}=\mathbf{0 . 0 8 3}{\text { bar } \mathrm{dm}^3}^{\mathrm{K}^{-1} \mathrm{~mol}^{-1}}\right)\).
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\(\mathrm{PV}=\mathrm{nRT}\)
or \(\mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{3.32 \mathrm{bar} \times 5 \mathrm{dm}^3}{4.0 \mathrm{~mol} \times 0.083 \mathrm{bar} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=50 \mathrm{~K}\)
or \(\mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{3.32 \mathrm{bar} \times 5 \mathrm{dm}^3}{4.0 \mathrm{~mol} \times 0.083 \mathrm{bar} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=50 \mathrm{~K}\)
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