On increasing the temperature of a gas filled in a close container by 1oC , its pressure increases by 0.4%, initial temperature of the gas is:

Question

On increasing the temperature of a gas filled in a close container by 1oC , its pressure increases by 0.4%, initial temperature of the gas is: (A)  25oC  (B)  250oC  (C)  250oK  (D)  2500oC

Tardigrade Admin · Accepted Answer

According to pressure law (P1/P2)=(T1/T2) ⇒ (P/P+(P× 0.4)100)=(x+273/x+274) (250/251)=(x+273/x+274) ∴ x=-23oC=250K

Q. On increasing the temperature of a gas filled in a close container by $1^{o} C$ , its pressure increases by 0.4%, initial temperature of the gas is:

$25^{o} C$

$250^{o} C$

$250^{o} K$

$2500^{o} C$

According to pressure law $\frac{P _{1}}{P _{2}} = \frac{T _{1}}{T _{2}}$ $\Rightarrow$ $\frac{P}{P + \frac{P \times 0.4}{100}} = \frac{x + 273}{x + 274}$ $\frac{250}{251} = \frac{x + 273}{x + 274}$ $∴$ $x = - 23^{o} C = 250 K$

Q. On increasing the temperature of a gas filled in a close container by 1oC , its pressure increases by 0.4%, initial temperature of the gas is:

25oC

250oC

250oK

2500oC