Question

17) It is found, on heating a gas, its volume increases by 50%
and pressure decreases to 60% of its original value. If the
original temperature was -15°C, find the temperature to
which it was heated ?
[Ans.-40.8°C]
​

Answer 1

Let initial volume be $\sf{V_1,}$ so the final volume will be,

$\longrightarrow\sf{V_2=V_1+V_1\cdot\dfrac{50}{100}}$

$\longrightarrow\sf{V_2=V_1+\dfrac{V_1}{2}}$

$\longrightarrow\sf{V_2=\dfrac{3V_1}{2}}$

Here the volume is increased by its 50%.

Let initial pressure be $\sf{P_1,}$ so the final pressure will be,

$\longrightarrow\sf{P_2=P_1\cdot\dfrac{60}{100}}$

$\longrightarrow\sf{P_2=\dfrac{3P_1}{5}}$

Here the pressure is decreased to its 60%.

The original temperature is,

$\longrightarrow\sf{T_1=-15^oC}$

$\longrightarrow\sf{T_1=(-15+273)\ K}$

$\longrightarrow\sf{T_1=258\ K}$

Let the final temperature to which the gas was heated be $\sf{T_2}$ in Kelvin.

We have the combined gas equation,

$\longrightarrow\sf{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}$

$\longrightarrow\sf{T_2=\dfrac{P_2V_2T_1}{P_1V_1}}$

$\longrightarrow\sf{T_2=\dfrac{\dfrac{3P_1}{5}\cdot\dfrac{3V_1}{2}\cdot258}{P_1V_1}}$

$\longrightarrow\sf{T_2=\dfrac{3}{5}\times\dfrac{3}{2}\times258}$

$\longrightarrow\sf{T_2=232.2\ K}$

$\longrightarrow\sf{T_2=(232.2-273)^oC}$

$\longrightarrow\sf{\underline{\underline{T_2=-40.8^oC}}}$

Hence the gas was heated to $\bf{-40.8^oC.}$

Answer 2

Question:

It is found, on heating a gas, its volume increases by 50% and pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated?

Answer:

V1=1

P1=1

T1=-15⁰C=258K

V2=1+50/100=3/2V1

P2=60/100=3/5P1

T2=?

Solution:

Gas Equation:

V1×P1/T1=V2×P2/T2

T2=(3P1/5×3V1/2×258)/P2V2

T2=3/5×3/2×258

T2=9/5×129

T2=1161/5

T2=(232.2-273)⁰C

T2=-40.8⁰C

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