How High Can Temperature Go with 10^4 Joules in a Monoatomic Gas Cylinder?

[shehry1]

Homework Statement

This is Pathria (2nd Ed) 1.6 and it seemed simple enough but the magnitude of the answer seems unbelievably large:

A cylindrical vessel 1 m long and .1 m in diameter is filled with a monoatomic gas at P = 1 atm and T = 300 K. The gas is heated by an electrical discharge along the axis of the vessel, which releases an energy of 10^4 joules. What will the temperature of the gas be immediately after the discharge.

Homework Equations

PV = nRT
E = 3nRT/2

The Attempt at a Solution

E2 - E1 = 10^4
nR = P1 * V/T1 where P1 = 1 and T1 = 300
V = pi * (.1/2)^2 * 1

E2 - E1 = (3/2)(nR)(T2 - T1)
10^4 = (3/2) (1 * V) (T2 - 300)/300

This seems to give an answer in the range of 10^8. Is this correct? Is it because the number of moles of the gas PV/RT are very few and that even this 'normal' increase in energy is leading to such epic temperatures. The magnitude is bothersome especially because even the sun's core is at a temperature of the order of 10^6

 

[Borek]

Check your units. Pay special attention to pressure.

 

[D H]

Check your units. Pay special attention to pressure.

... and to the gas constant R.

 

[Borek]

There is no R in the final equation

 

[Andrew Mason]

Homework Statement

This is Pathria (2nd Ed) 1.6 and it seemed simple enough but the magnitude of the answer seems unbelievably large:

A cylindrical vessel 1 m long and .1 m in diameter is filled with a monoatomic gas at P = 1 atm and T = 300 K. The gas is heated by an electrical discharge along the axis of the vessel, which releases an energy of 10^4 joules. What will the temperature of the gas be immediately after the discharge.

Homework Equations

PV = nRT
E = 3nRT/2

The Attempt at a Solution

E2 - E1 = 10^4
nR = P1 * V/T1 where P1 = 1 and T1 = 300
V = pi * (.1/2)^2 * 1

E2 - E1 = (3/2)(nR)(T2 - T1)
nR = P_1V/T_1

This is correct. You have to be clearer in your reasoning. Isolate T2, find nR and V and plug in the numbers.

Another way of approaching this is to work out the heat capacity of the gas nC_V. Since the volume does not change, the change in temperature is just the added heat energy divided by the heat capacity:

$$nC_v = \frac{3}{2}nR = \frac{3PV}{2T} = 3\times 100325\times 7.85\times 10^{-3}/2 \times 300 = 3.93 J/K$$

$$T_2 = \Delta E/nC_v + T_1$$

AM

 

[D H]

There is no R in the final equation

Point taken. What Borek was talking about in post #2 is this:

10^4 = (3/2) (1 * V) (T2 - 300)/300

shehry1, you are being very sloppy with units here (there are no units here!) That's a freshman mistake. From your other posts (Jackson, spin matrices, theoretical mechanics), you are not a freshman.