Question

2 moles of monoatomic ideal gas has temperature \(T\) and 6 moles of monoatomic ideal gas has temperature \(2T\). Find the temperature of mixture.

• A. \(\dfrac{7T}{4}\)
• B. \(\dfrac{3T}{4}\)
• C. \(\dfrac{4T}{4}\)
• D. \(7T\)

Hint:

When two ideal gases of same type are mixed, final temperature can be found by equating total initial internal energy with total final internal energyFor monoatomic gas, always use \(U=\frac{3}{2}nRT\).

Answer 1

The Correct Option is A

Step 1: Use conservation of internal energy.
Since both gases are monoatomic ideal gases, their molar heat capacity at constant volume is same.
When two samples of the same ideal gas are mixed without heat loss, final temperature is obtained from energy balance.
For monoatomic ideal gas, internal energy is:
\[ U=\frac{3}{2}nRT \]

Step 2: Write internal energy of both gases before mixing.
For first gas:
\[ U_1=\frac{3}{2}(B)RT \]
\[ U_1=3RT \]
For second gas:
\[ U_2=\frac{3}{2}(6)R(2T) \]
\[ U_2=18RT \]
So, total initial internal energy is:
\[ U_{\text{initial}}=U_1+U_2=3RT+18RT=21RT \]

Step 3: Write internal energy after mixing.
Total number of moles after mixing:
\[ n_{\text{total}}=2+6=8 \]
Let final temperature of mixture be \(T_f\). Then,
\[ U_{\text{final}}=\frac{3}{2}(8)RT_f \]
\[ U_{\text{final}}=12RT_f \]

Step 4: Apply conservation of energy.
Since no heat is lost:
\[ U_{\text{initial}}=U_{\text{final}} \]
\[ 21RT=12RT_f \]
\[ T_f=\frac{21T}{12} \]
\[ T_f=\frac{7T}{4} \]

Step 5: Match with the options.
Hence, the temperature of the mixture is:
\[ \boxed{\frac{7T}{4}} \]
Final Answer:
\[ \boxed{\frac{7T}{4}} \]