JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 15)
Two flasks I and II shown below are connected by a valve of negligible volume.
_27th_August_Evening_Shift_en_15_1.png)
When the valve is opened, the final pressure of the system in bar is x $$\times$$ 10$$-$$2. The value of x is __________. (Integer answer)
[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$$-$$1; R = 8.31 J mol$$-$$1 K$$-$$1]
When the valve is opened, the final pressure of the system in bar is x $$\times$$ 10$$-$$2. The value of x is __________. (Integer answer)
[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$$-$$1; R = 8.31 J mol$$-$$1 K$$-$$1]
Answer
84
Explanation
Applying; (nI + nII)initial = (nI + nII)final
$$\Rightarrow$$ Assuming the system attains a final temperature of T (such that 300 < T < 60)
$$\Rightarrow$$ $$\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$$
$$\Rightarrow$$ n1Cm(300 $$-$$ T) = nIICm(T $$-$$ 60)
$$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$$
$$\Rightarrow$$ 14(300 $$-$$ T) = T $$-$$ 60
$$\Rightarrow$$ $${{(14 \times 300 + 60)} \over {15}} = T$$
$$\Rightarrow$$ T = 284 K (final temperature)
$$\Rightarrow$$ If the final pressure = P
$$\Rightarrow$$ (nI + nII)final = $$\left( {{{3.0} \over {28}}} \right)$$
$$\Rightarrow$$$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$$
$$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$$
$$\Rightarrow$$ 0.84287 bar
$$\Rightarrow$$ 84.28 $$\times$$ 10$$-$$2 bar
$$\Rightarrow$$ 84
$$\Rightarrow$$ Assuming the system attains a final temperature of T (such that 300 < T < 60)
$$\Rightarrow$$ $$\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$$
$$\Rightarrow$$ n1Cm(300 $$-$$ T) = nIICm(T $$-$$ 60)
$$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$$
$$\Rightarrow$$ 14(300 $$-$$ T) = T $$-$$ 60
$$\Rightarrow$$ $${{(14 \times 300 + 60)} \over {15}} = T$$
$$\Rightarrow$$ T = 284 K (final temperature)
$$\Rightarrow$$ If the final pressure = P
$$\Rightarrow$$ (nI + nII)final = $$\left( {{{3.0} \over {28}}} \right)$$
$$\Rightarrow$$$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$$
$$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$$
$$\Rightarrow$$ 0.84287 bar
$$\Rightarrow$$ 84.28 $$\times$$ 10$$-$$2 bar
$$\Rightarrow$$ 84
Comments (0)
