Question

The molecular weights of $O_2$ and $N_2$ are 32 and 28 respectively. At ${15}^{o}C$, the pressure of 1 gm $O_2$ will be the same as that of 1 gm $N_2$ in the same bottle at the temperature :

• A. $-{21}^{o}C$
• B. ${13}^{o}C$
• C. ${15}^{o}C$
• D. ${56.4}^{o}C$

Answer 1

Answer: (A)

$-{21}^{o}C$

Volume is the same.

For 1g O${}_2$ :

T = 15${}^o$C = 273 + 15 = 288K

n = $\frac{1}{32}$

PV = nRT

PV = $\frac{1}{32}$ x R x 288 (i)

For 1g N${}_2$ :

n = $\frac{1}{28}$

PV = nRT

PV = $\frac{1}{28}$ x R x T (ii)

Equating both equation.

$\frac{1}{32}$ x R x 288 = $\frac{1}{28}$ x R x T

T = $\frac{288\times 28}{32}$

T = 252K

T = -21${}^o$C