3 moles of an ideal monoatomic gas is heated at constant pressure of one atmosphere from 0- C to 70- C- Then- work done by the gas isA- 1-74 kJB- 0-5 kJC- 6-98 kJD- 11-3 kJ

The correct option is A 1.74 kJFor monoatomic gas, degrees of freedom is 3 Thus, C_v=3/2R and C_p=5/2R R=8.31 J/mol K dQ=dU+W dQ=nC_pdT=3×5/2×8.31×(70 0) dQ=43 ...

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Byju's Answer

Standard XII

Chemistry

Entropy Change in Isochoric Process

3 moles of an...

Question

3 moles of an ideal monoatomic gas is heated at constant pressure of one atmosphere from $0^{\circ}$ C to $70^{\circ}$ C. Then, work done by the gas is

1.74

k J

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0.5

k J

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6.98

k J

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11.3

k J

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Solution

The correct option is A 1.74 $k J$
For monoatomic gas, degrees of freedom is 3
Thus, $C_{v} = \frac{3}{2} R$ and $C_{p} = \frac{5}{2} R$
$R = 8.31 J / m o l - K$
$d Q = d U + W$
$d Q = n C_{p} d T = 3 \times \frac{5}{2} \times 8.31 \times (70 - 0)$
$d Q = 4362.75 J$
$d U = n C_{v} d T = 3 \times \frac{3}{2} \times 8.31 \times (70 - 0)$
$d U = 2617.65 J$
$W = 4362.75 - 2617.65 = 1745.1 J$
$W = 1.74 K J$

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3 moles of an ideal monoatomic gas is heated at constant pressure of one atmosphere from 0∘C to 70∘C. Then, work done by the gas is

The correct option is A 1.74 kJFor monoatomic gas, degrees of freedom is 3 Thus, Cv=32R and Cp=52R R=8.31 J/mol−K dQ=dU+W dQ=nCpdT=3×52×8.31×(70−0) dQ=4362.75 J dU=nCvdT=3×32×8.31×(70−0) dU=2617.65 J W=4362.75−2617.65=1745.1 J W=1.74 KJ

3 moles of an ideal monoatomic gas is heated at constant pressure of one atmosphere from $0^{\circ}$ C to $70^{\circ}$ C. Then, work done by the gas is