Question

If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,

Answer 1

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

$\frac{P1V1}{RT1}$ = $\frac{P2V2}{RT2}$

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

$\frac{1*20}{R*10} = \frac{10*V2}{R*100}$

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

$\frac{1*20*100}{10*10} = V2$

∴ V2 = $\frac{2000}{100}$

∴ V2 = 20mL