# The amount of heat energy required to raise the temperature of 1g of helium at NTP from T1 K to T2 K is?

The amount of heat energy required to raise the temperature of 1g of helium at NTP from $$T_1 \pu{K}$$ to $$T_2 \pu{K}$$ is?

My confusion is, that as the gas is at NTP, therefore, it is an isochoric process as well as isometric process.

• There are at least two possible answers to this question. It depends on the conditions of the heating process. The gas can be heated at constant volume or at constant pressure. In the first case, Cv = 3R/2 must be used. In the second case, Cp = 5R/2 must be used. – Maurice Apr 7 '20 at 9:04
• Just tell me if this is right – Yash Apr 7 '20 at 15:22
• A gas is at constant pressure if at NTP as NTP means 1 atm – Yash Apr 7 '20 at 15:22
• Also at constant volume because NTP means 22.4 l for 1 mole – Yash Apr 7 '20 at 15:23
• Also pls look into this working if we consider both isobaric and isochoric then delta U=delta Q -delta W so change in work is zero so this should be U=Q which means – Yash Apr 7 '20 at 15:25

Well. A gas is at constant pressure if the pressure does not change in the container with the time. This pressure may be 1 atm, but it may have any other value, provided this value does not change during your experiment or your measurement.

A gas at constant volume is a gas inside a cloud vessel, where the temperature or itse composition may change. This property has nothing to do with the volume 22,4 L/mol at STP.

Now, if the volume of a gas is constant, there is no work done on or by this gas, and Delta U = Delta Q

• What is cloud vessel? – TheLearner Apr 7 '20 at 16:34
• The pressure is 1 atm because it is at NTP if this isn't the idea then why NTP is used in the question and what does it mean ? – Yash Apr 7 '20 at 17:41

From thermodynamics,

$$\Delta U = Q - W$$

$$\Delta U = Q - \Delta(PV)$$

$$\Delta(PV) + \Delta U = Q - \Delta(PV) + \Delta(PV)$$

We know that, $$\Delta(PV) + \Delta U = \Delta H$$

$$\therefore$$ assuming constant pressure, $$\Delta P = 0$$,

$$\Delta H = Q$$

$$mC_p\Delta T = Q$$

From a differential perspective, for the case of a gas at NTP (1 atm), we can use the definition of $$Q=m_{g}\int^{T_2}_{T_1}C_p(T)dT$$. Where $$m_g$$ is the mass of gas, $$C_p(T)$$ is the heat capacity (at constant pressure) of the gas at a given temperature (heat capacities for gases are widely reported in by institutions like NIST); generally in a form like $$C_p(T) = a + bT + cT^2 + dT^3 ...$$