Question

Calculate the final volume of a gas, if the pressure of the gas, originally at s.t.p. is doubled and its temperature is tripled.

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Answer 1

Answer:

ANSWER

Suppose the volume at STP to be V and pressure to be P and temperature to be T

Now, the number of moles of the gas X present

  n=RTPV

 PV=nRT

Now, the second condition

  P2=2P

 T2=3T

Now, the volume

V2=P2nRT2

Now, substituting the value of T2 and P2

  V2=P2nRT2

 V2=TPV×2P3T

 V2=23V

 V2=1.5V

So, the final volume will be 1.5 fold the original volume.

Explanation:

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Answer 2

$\huge{ \underline{ \bf{Given \: : }}}$

Initial Conditions [s.t.p]

• $\tt \: p_1 = 760mm. \: of \: hg$
• $\tt \: v_1 = v_1 \: lits$
• $\tt \: t_1 = 273k$

Final Conditions

• $\tt \: p_2 = 2 \times 760mm.$
• $\tt \: v_2 = ? \: lits$
• $\tt \: t_2 = 3 \times 273k$

$\huge{ \underline{ \underline{ \bf{Solution \: : - }}}}$

Using gas equation

$\color{purple} \leadsto \red{ \underline{ \underline{ \boxed{ \green{ \tt{ \frac{p _1v _ 1}{t _1} = \frac{p _2v _2 }{t _2 } }}}}}}$

Substituting the values

$\tt \to \frac{760 \times v _1}{273} = \frac{2 \times 760 \times v_2}{ 3 \times 273}$

$\tt \to v_2 = \frac{ \cancel{760} \times v_1 \times3 \times \cancel{273}}{2 \times \cancel{760} \times \cancel{273}}$

$\tt \to \: v_2 = 1.5v_1$

Therefore, the final volume is 1 1/2 times the volume of the gas originally at s.t.p

$\huge{ \underline{ \bf{ \pink\star \: Know \: more \: \pink \star}}}$

The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions, although it has several limitations.