Question

Two monoatomic ideal gases A and B occupying the same volume V are at the same temperature T and pressure p. U they are mixed, the resultant mixture has volume V and temperature T. The pressure of the mixture is

• A. $p$
• B. $\frac{p}{2}$
• C. $4p$
• D. $2p$

Answer 1

Answer: (D)

$2p$

$\frac{p_1{V}_1}{T_1}=\frac{p_{mix}\times V_{mix}}{T_{mix}}$
$T_1=T_{mix}=T$
$V_1=V+V$
$V_{mix}=V$
$=\frac{p\times 2V}{T}=\frac{p_{mix}}{T}\times V$
$p_{mix}=2p$