Question

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is removed from the tank. What is the pressure of the remaining gas in the tank?

Answer 1

The pressure of the remaining gas in the tank is 6.4 atm.

Step-by-step explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

`PV=nRT`

For a gas

`P_(1)V_(1)=nRT_(1)`

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

`10* V=nR*286`....(I)

When the temperature of the gas is increased

Then,

`P_(2)V_(2)=(n)/(2)RT_(2)`....(II)

Divided equation (I) by equation (II)

`(P_(1)V)/(P_(2)V)=(nRT_(1))/((n)/(2)RT_(2))`

`(10* V)/(P_(2)V)=(nR*286)/((n)/(2)R368)`

`P_(2)=(10*368)/(2*286)`

`P_(2)= 6.433\ atm`

`P_(2)=6.4\ atm`

Hence, The pressure of the remaining gas in the tank is 6.4 atm.