A cylinder contains 10 kg of gas at pressure of 107N-m2- When final pressure is reduce to 2-5- 106N-m2 then quantity of gas taken out of the cylinder will be temperature of gas is constant-

The correct option is D 7.5 kg The correct answer is option(A). PV= mMRT m∝ P Hence, m_210=2.5× 10^610^7 m_2=2.5Kg Δ m=10 2.5=7.5Kg ...

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Standard X

Physics

Ideal Gas Equation

A cylinder co...

Question

A cylinder contains $10 k g$ of gas at pressure of $10^{7} N / m^{2}$ . When final pressure is reduce to $2.5 \times 10^{6} N / m^{2}$ then quantity of gas taken out of the cylinder will be (temperature of gas is constant).

15.2 k g

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3.7 k g

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Zero

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7.5 k g

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Solution

The correct option is D $7.5 k g$
The correct answer is option(A).
$P V = \frac{m}{M} R T$
$⟹ m \propto P$
Hence, $\frac{m_{2}}{10} = \frac{2.5 \times 10^{6}}{10^{7}}$
$⟹ m_{2} = 2.5 K g$
$Δ m = 10 - 2.5 = 7.5 K g$ is the quantity of gas taken out o the cylinder.

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A cylinder contains 10 kg of gas at pressure of 107N/m2. When final pressure is reduce to 2.5×106N/m2 then quantity of gas taken out of the cylinder will be (temperature of gas is constant).

The correct option is D 7.5 kgThe correct answer is option(A).PV=mMRT⟹m∝PHence, m210=2.5×106107⟹m2=2.5KgΔm=10−2.5=7.5Kg is the quantity of gas taken out o the cylinder.

A cylinder contains $10 k g$ of gas at pressure of $10^{7} N / m^{2}$ . When final pressure is reduce to $2.5 \times 10^{6} N / m^{2}$ then quantity of gas taken out of the cylinder will be (temperature of gas is constant).

The correct option is D $7.5 k g$
The correct answer is option(A).
$P V = \frac{m}{M} R T$
$⟹ m \propto P$
Hence, $\frac{m_{2}}{10} = \frac{2.5 \times 10^{6}}{10^{7}}$
$⟹ m_{2} = 2.5 K g$
$Δ m = 10 - 2.5 = 7.5 K g$ is the quantity of gas taken out o the cylinder.