Question

Solve this..

Kota test paper question..​

Answer 1

The pressure in both the bulbs before placing in ice and hot bath is the same and is taken as,

• $\sf{P_{1i}=P_{2i}=P}$

Since both were at STP initially, the temperature is same in both bulbs before placing, thus,

• $\sf{T_{1i}=T_{2i}=273\ K}$

After placing in ice and hot bath, the pressure in each bulb becomes 1.5 times. So,

• $\sf{P_{1f}=P_{2f}=1.5\,P}$

The final temperature in the bulb placed in ice is the same as the temperature of ice (melting point of ice), i.e.,

• $\sf{T_{1f}=273\ K}$

Let the final temperature of the bulb in hot bath (temperature of hot bath) be $\sf{T_{2f}.}$

Then by Gay Lussac's law, since the volume is constant here,

$\displaystyle\longrightarrow\sf{\dfrac{P_{1i}}{T_{1i}}+\dfrac{P_{2i}}{T_{2i}}=\dfrac{P_{1f}}{T_{1f}}+\dfrac{P_{2f}}{T_{2f}}}$

$\displaystyle\longrightarrow\sf{\dfrac{P}{273}+\dfrac{P}{273}=\dfrac{1.5\,P}{273}+\dfrac{1.5\,P}{T_{2f}}}$

$\displaystyle\longrightarrow\sf{\dfrac{2}{273}=\dfrac{1.5}{273}+\dfrac{1.5}{T_{2f}}}$

$\displaystyle\longrightarrow\sf{\dfrac{1.5}{T_{2f}}=\dfrac{0.5}{273}}$

$\displaystyle\longrightarrow\sf{T_{2f}=3\times273\ K}$

$\displaystyle\longrightarrow\sf{T_{2f}=819\ K}$

$\displaystyle\longrightarrow\sf{\underline{\underline{T_{2f}=546^oC}}}$

Answer 2

Quantity of gas in these bulbs is constant i.e,

$\huge\boxed{\text{\fcolorbox{Initial no of moles = final no of moles}}}$

⇒ n₁ + n₂ = n₁ + n₂

$\frac{PV}{R(273)} + \frac{PV}{R(273)} = \frac{1.5PV}{R(273)} + \frac{1.5PV}{R(273)}$

$\large{\green{\boxed{\frac{2}{273}=\frac{1.5}{273}+\frac{1.5}{T}}}}}}}}}}}}}}}}}$

​⇒ T = 819 K

= 546° C