Question

It $2$ moles of an ideal monoatomic gas at temperature $T_{0}$ is mixed with $4$ moles of another ideal monoatomic gas at temperature $2 T_{0}$, then the temperature of the mixture is:

• A. $\frac{5}{3} T_{0}$
• B. $\frac{3}{2} T_{0}$
• C. $\frac{4}{3} T_{0}$
• D. $\frac{5}{4} T_{0}$

Answer 1

Answer: (A)

The pressure will increase
$p_{1}+p_{2}=p$
From ideal gas equation,
$p v=n R T$
$p=\frac{n R T}{v}$
Volume remains same
$\frac{2 R T_{0}}{v}+\frac{4 R\left(2 T_{0}\right)}{v}=\frac{6 R T}{v}$
where $T=$ temperature of mixture
$10\, T_{0} =6\, T$
$T =\frac{5 T_{0}}{3}$