A 8-3 L cylinder contains 128 g of O2 gas at 27 - C- What is the mass of O2 gas mustb e released to reduce the pressure to 6-0 bar-

We know, p_1V_1=n_1RT_1 ⇒ P_1=n_1RT_1V_1=w_1RT_1m_1V_1=128× 0.083× 30032× 8.3 = 12 bar Since the volume is constant, at constant temperature, P_2P_1=n_2n_1 612 ...

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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A 8.3 L cylin...

Question

A $8.3 L$ cylinder contains $128 g$ of $O_{2}$ gas at $27^{\circ} C$ . What is the mass of $O_{2}$ gas mustb e released to reduce the pressure to $6.0 bar$ ?

32.0 g

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64.0 g

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16.0 g

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8.0 g

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Solution

The correct option is B $64.0 g$
We know,
$p_{1} V_{1} = n_{1} R T_{1}$
$\Rightarrow P_{1} = \frac{n_{1} R T_{1}}{V_{1}} = \frac{w_{1} R T_{1}}{m_{1} V_{1}} = \frac{128 \times 0.083 \times 300}{32 \times 8.3}$
= 12 bar

Since the volume is constant, at constant temperature,
$\frac{P_{2}}{P_{1}} = \frac{n_{2}}{n_{1}}$
$\frac{6}{12} = \frac{n_{2}}{4}$
Hence, $n_{2} = 2 moles$
Thus, mass of $O_{2}$ released $= 2 \times 32 g = 64 g$

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A 8.3 L cylinder contains 128 g of O2 gas at 27 ∘C. What is the mass of O2 gas mustb e released to reduce the pressure to 6.0 bar?

The correct option is B 64.0 gWe know, p1V1=n1RT1 ⇒P1=n1RT1V1=w1RT1m1V1=128×0.083×30032×8.3 = 12 bar Since the volume is constant, at constant temperature, P2P1=n2n1 612=n24 Hence, n2=2 moles Thus, mass of O2 released =2×32 g=64 g

A $8.3 L$ cylinder contains $128 g$ of $O_{2}$ gas at $27^{\circ} C$ . What is the mass of $O_{2}$ gas mustb e released to reduce the pressure to $6.0 bar$ ?