Three moles of hydrogen gas arc compressed isothermally and reversibly from 60 L to 20 L $8.50 k J$ of work is done on it. Assuming ideal behavior, calculate the temperature of the.gas. Since the gas behaves ideally, reversible work of compression is given by
$w = 2.30 n R T l o g \frac{V_{2}}{V_{1}}$
Here $n = 3 m o l, V_{1} = 60 L, V_{2} = 20 L, w = 8.5 \times 10^{3} J$
$8.50 \times 10^{3} J = 2.303 \times (3) (8.314) T$ log $\frac{60}{20}$

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Solution

Given: $n = 3, N_{1} = 60 L, V_{2} = 20 L, W = 2.50 K J = 2500 J$
Isothermal and reversible compression
$W = 2.303 R T l o g \frac{V_{1}}{V_{2}}$
$2500 J = 2.303 \times 8.314 \times T l o g \frac{60}{20}$
$T = \frac{2500 J}{2.303 \times 8.314 \times 0.4771}$
$T = 930 K$

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Given: n=3, N1=60 L, V2=20 L, W=2.50 KJ=2500JIsothermal and reversible compressionW=2.303 RT log V1V22500 J=2.303×8.314×T log 6020T=2500 J2.303×8.314×0.4771T=930 K

Given: $n = 3, N_{1} = 60 L, V_{2} = 20 L, W = 2.50 K J = 2500 J$
Isothermal and reversible compression
$W = 2.303 R T l o g \frac{V_{1}}{V_{2}}$
$2500 J = 2.303 \times 8.314 \times T l o g \frac{60}{20}$
$T = \frac{2500 J}{2.303 \times 8.314 \times 0.4771}$
$T = 930 K$