Bright Tutorials Bright Tutorials
ICSE Class 9 Chemistry Question 6 of 29

Study of Gas Laws — Question 14

Back to all questions
14
Question

Question 14

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

Initial conditions [S.T.P.] :

P1 = Initial pressure of the gas = P
V1 = Initial volume of the gas = V
T1 = Initial temperature of the gas = -15°C = -15 + 273 = 258 K

Final conditions :

P2 (Final pressure) = pressure decreases to 60% of its original value

= 60100\dfrac{60}{100} of P

= 35\dfrac{3}{5}P

V2 (Final volume) = volume increases by 50% of its original value

= 1 + 50100\dfrac{50}{100} of V

= 100+50100\dfrac{100 + 50}{100} of V

= 150100\dfrac{150}{100} of V

= 32\dfrac{3}{2}V

T2 (Final temperature) = ?

By Gas Law:

P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}

Substituting the values :

PV 258=35P×32VT2T2=3×3×2585×2=232210T2=232.2K\dfrac{\text{PV }}{258} = \dfrac{\dfrac{3}{5} \text{P}\times\dfrac{3}{2}\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{3\times 3\times 258}{5 \times 2} = \dfrac{2322}{10} \\[1em] \text{T}_2 = 232.2 \text{K} \\[1em]

∴ Final temperature of the gas = 232.2 - 273 = -40.8°C