The pressure exerted by 12 g of an ideal gas at t- C in a vessel of V L volume is 1 atm- When the temperature is increased by 10- C in the same vessel the pressure increases by 10 - The value of t is -A- 173- CB- 173 kC- 100- CD- 100 k

The correct option is D Apply Gay – lussac’s law; P_1/T_1=P_2/T_2 P_1 = 1 atm P_2 = 1+ ( 10/100× 1 )=110/100=1.1 atm T_1 = (t+273)K T_2=(t+283)K 1/(t+2 ...

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Byju's Answer

Standard XII

Physics

Bulk Modulus

The pressure ...

Question

The pressure exerted by 12g of an ideal gas at $t^{\circ} C$ in a vessel of V L volume is 1 atm. When the temperature is increased by $10^{\circ} C$ in the same vessel the pressure increases by $10 %$ . The value of t is –

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Solution

The correct option is D

Apply Gay – lussac’s law; $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

$P_{1} = 1 a t m P_{2} = 1 + (\frac{10}{100} \times 1) = \frac{110}{100} = 1.1 a t m T_{1} = (t + 273) K T_{2} = (t + 283) K \frac{1}{(t + 273)} = \frac{1.1}{(t + 283)}$

1.1t + 300.3 = t + 283

0.01t = 17.3

$t = - 173^{\circ} C$

$∴$ t = 173 + 273 = 100 K

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The pressure exerted by 12g of an ideal gas at t∘C in a vessel of V L volume is 1 atm. When the temperature is increased by 10∘C in the same vessel the pressure increases by 10%. The value of t is –

The correct option is D Apply Gay – lussac’s law; P1T1=P2T2 P1=1 atm P2=1+(10100×1)=110100=1.1 atmT1=(t+273)K T2=(t+283)K1(t+273)=1.1(t+283) 1.1t + 300.3 = t + 283 0.01t = 17.3 t=−173∘C ∴ t = 173 + 273 = 100 K

The pressure exerted by 12g of an ideal gas at $t^{\circ} C$ in a vessel of V L volume is 1 atm. When the temperature is increased by $10^{\circ} C$ in the same vessel the pressure increases by $10 %$ . The value of t is –