The total energy of 1 mole of an ideal monatomic gas 27 -circ- C is - text-cal-

Question

Toppr Admin · Accepted Answer

As we know-KE per mole -32RT-32-xD7-2-xD7-300-900calWhere R- 1-99 cal K-x2212-1 mol-x2212-1The total energy of 1 mole of an ideal monatomic gas at 27 -x2218-C is900 cal-

The total energy of 1 mole of an ideal monatomic gas at $27^{\circ} C$ is ............ $cal$ .

As we know,
$K E$ per mole $= \frac{3}{2} R T = \frac{3}{2} \times 2 \times 300 = 900 c a l$

Where R= 1.99 cal K $^{- 1}$ mol $^{- 1}$

The total energy of 1 mole of an ideal monatomic gas at 27 $^{\circ}$ C is900 cal.

The total energy of 1 mole of an ideal monatomic gas at 27 ∘C is ............ cal.

As we know,KE per mole =32RT=32×2×300=900calWhere R= 1.99 cal K−1 mol−1The total energy of 1 mole of an ideal monatomic gas at 27 ∘C is900 cal.

The total energy of 1 mole of an ideal monatomic gas at $27^{\circ} C$ is ............ $cal$ .

As we know,
$K E$ per mole $= \frac{3}{2} R T = \frac{3}{2} \times 2 \times 300 = 900 c a l$

Where R= 1.99 cal K $^{- 1}$ mol $^{- 1}$

The total energy of 1 mole of an ideal monatomic gas at 27 $^{\circ}$ C is900 cal.