3 moles of an ideal gas at a temperature of 27° C are mixed with 2 moles of an ideal gas at a temperature 227° C, determine the equilibrium temperature of the mixture, assuming no loss of energy

Question

3 moles of an ideal gas at a temperature of 27° C are mixed with 2 moles of an ideal gas at a temperature 227° C, determine the equilibrium temperature of the mixture, assuming no loss of energy (A) 327° C (B) 107° C (C) 318° C (D) 410° C

Tardigrade Admin · Accepted Answer

Energy possessed by the ideal gas at 27° C is E1=3((3/2) R × 300) =((2700 R/2)) Energy possessed by the ideal gas at 227° C is E2=2((3 R/2) × 500)=1500 R If T be the equilibrium temperature, of the mixture, then its energy will be E2=5((3 R T/2)) Since, energy remains conserved, Em=E1+E2 or 5((3 R T/2))=(2700 R/2)+1500 R or T=380 K or 107° C

Q. $3$ moles of an ideal gas at a temperature of $2 7^{\circ} C$ are mixed with $2$ moles of an ideal gas at a temperature $22 7^{\circ} C$ , determine the equilibrium temperature of the mixture, assuming no loss of energy

$32 7^{\circ} C$

$10 7^{\circ} C$

$31 8^{\circ} C$

$41 0^{\circ} C$

Q. 3 moles of an ideal gas at a temperature of 27∘C are mixed with 2 moles of an ideal gas at a temperature 227∘C, determine the equilibrium temperature of the mixture, assuming no loss of energy

327∘C

107∘C

318∘C

410∘C

Q. $3$ moles of an ideal gas at a temperature of $2 7^{\circ} C$ are mixed with $2$ moles of an ideal gas at a temperature $22 7^{\circ} C$ , determine the equilibrium temperature of the mixture, assuming no loss of energy

$32 7^{\circ} C$

$10 7^{\circ} C$

$31 8^{\circ} C$

$41 0^{\circ} C$